Projectile Motion Problems with Step-by-Step Solutions
Learn how to solve projectile motion problems in physics with step-by-step solutions. This tutorial covers height, time, maximum height, and the time an object hits the ground. Perfect for high school and college students practicing physics and math problems.
Table of Contents
Review of Projectile Motion Equations
If air resistance is ignored, the height \( h \) of a projectile above the ground after \( t \) seconds is given by the quadratic kinematic equation:
\[ h(t) = -\frac{1}{2} g t^{2} + V_{0} t + h_{0} \]where:
- \( g \) is the acceleration due to gravity: approximately \( 32 \,\text{ft/s}^{2} \) or \( 9.8 \,\text{m/s}^{2} \) on Earth,
- \( V_{0} \) is the initial velocity at \( t = 0 \),
- \( h_{0} \) is the initial height at \( t = 0 \).
For an object that is simply dropped (initial velocity is zero), the distance traveled downwards is:
\[ d(t) = \frac{1}{2} g t^2 \]Because the height formula forms a quadratic function of the form \( h(t) = a t^2 + b t + c \) with a negative coefficient \( a \), the maximum height occurs at the vertex of the parabola:
\[ t = - \frac{b}{2 a} \]Problem 1: Projectile from a Building
The formula \( h(t) = -16 t^2 + 32 t + 80 \) gives the height \( h \) above ground, in feet, of an object thrown, at \( t = 0 \), straight upward from the top of an 80-foot building.
- What is the highest point reached by the object?
- How long does it take the object to reach its highest point?
- After how many seconds does the object hit the ground?
- For how many seconds is the height of the object higher than 90 feet?
View Step-by-Step Solution
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Highest Point: The height \( h \) given above is a quadratic function. The graph of \( h \) as a function of time \( t \) forms a parabola opening downwards, and the maximum height occurs at the vertex. For \( h(t) = a t^{2} + b t + c \), the vertex is located at \( t = \frac{-b}{2a} \).
Substitute the values: \[ t = \frac{-32}{2(-16)} = 1 \text{ second} \] Substitute \( t = 1 \) back into the height equation to find the maximum height: \[ h(1) = -16(1)^{2} + 32(1) + 80 = 96 \text{ feet} \]Answer (a): The highest point is \( \boxed{96 \text{ feet}} \). -
Time to Highest Point: As calculated in part (a), it takes exactly 1 second for the object to reach its highest point.
Answer (b): \( \boxed{t = 1 \text{ second}} \).
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Time to Hit Ground: At the ground, \( h = 0 \). Solving the equation \( h(t) = 0 \) gives the time \( t \) at which the object hits the ground:
\[ -16t^{2} + 32t + 80 = 0 \]
Divide the entire equation by -16 to simplify:
\[ t^2 - 2t - 5 = 0 \]
Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-5)}}{2} = \frac{2 \pm \sqrt{24}}{2} = 1 \pm \sqrt{6} \]
Since time must be positive, we take \( 1 + \sqrt{6} \approx 3.45 \) seconds. So, it takes approximately 3.5 seconds for the object to hit the ground.
Answer (c): \( \boxed{t = 1 + \sqrt{6} \approx 3.45 \text{ seconds}} \).
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Time Above 90 Feet: The object is higher than 90 feet for all values of \( t \) satisfying the inequality \( h(t) \gt 90 \):
\[ -16t^{2} + 32t + 80 \gt 90 \]
\[ -16t^{2} + 32t - 10 \gt 0 \]
Multiply by \(-1\) and reverse the inequality:
\[ 16t^{2} - 32t + 10 \lt 0 \]
Using the quadratic formula to find the roots of \( 16t^2 - 32t + 10 = 0 \):
\[ t = \frac{32 \pm \sqrt{1024 - 640}}{32} = \frac{32 \pm \sqrt{384}}{32} = \frac{32 \pm 8\sqrt{6}}{32} = 1 \pm \frac{\sqrt{6}}{4} \]
Numerically, \( t \approx 0.3876 \) and \( t \approx 1.6124 \). The inequality is satisfied between these two times:
\[ 0.39 \lt t \lt 1.61 \text{ seconds} \]
The total duration the object is higher than 90 feet is the difference between these times:
\[ 1.61 - 0.39 \approx 1.22 \text{ seconds} \]
Answer (d): \( \boxed{\approx 1.22 \text{ seconds}} \).
Problem 2: Rock Dropped into a Well
A rock is dropped into a well and the distance traveled downwards is \( d = 16t^{2} \) (in feet), where \( t \) is the time in seconds. If the water splash is heard exactly 3 seconds after the rock was dropped, and the speed of sound is \( 1100 \text{ ft/sec} \), approximate the depth of the well.
View Step-by-Step Solution
Let \( T_{1} \) be the time it takes the rock to reach the bottom of the well. If \( H \) is the depth of the well, then the distance fallen is:
\[ H = 16 T_{1}^{2} \]Let \( T_{2} \) be the time it takes the sound wave to travel from the bottom back to the top of the well. Traveling at constant speed:
\[ H = 1100 T_{2} \]The total time observed is the sum of the falling time and the sound travel time:
\[ T_{1} + T_{2} = 3 \implies T_{2} = 3 - T_{1} \]Set the two expressions for \( H \) equal to each other to eliminate \( H \):
\[ 16 T_{1}^{2} = 1100 T_{2} \] \[ 16 T_{1}^{2} = 1100 (3 - T_{1}) \]Expand and write as a standard quadratic equation:
\[ 16 T_{1}^{2} + 1100 T_{1} - 3300 = 0 \]Divide by 4 to simplify:
\[ 4 T_{1}^{2} + 275 T_{1} - 825 = 0 \]Solve for \( T_{1} \) using the quadratic formula:
\[ T_{1} = \frac{-275 \pm \sqrt{275^{2} + 4(4)(825)}}{2(4)} = \frac{-275 \pm \sqrt{75625 + 13200}}{8} = \frac{-275 \pm \sqrt{88825}}{8} \]Numerically, \( \sqrt{88825} \approx 298.04 \), so:
\[ T_{1} \approx \frac{-275 + 298.04}{8} \approx \frac{23.04}{8} \approx 2.88 \text{ seconds} \]Now, substitute \( T_{1} \) back into the distance formula to calculate the depth of the well:
\[ H = 16 T_{1}^{2} = 16 (2.88)^{2} \approx 16(8.2944) \approx 132.7 \text{ feet} \]Verification: Sound travel time \( T_2 = 3 - 2.88 = 0.12 \) s, so distance \( = 1100 \times 0.12 = 132 \) ft, consistent with the falling distance. ✓
Problem 3: Object Thrown from Building
From the top of a building, an object is thrown upward with an initial speed of \( 64 \text{ ft/sec} \). It touches the ground 5 seconds later. What is the height of the building?
View Step-by-Step Solution
The formula for the height \( h \) of a projectile thrown upward is given by:
\[ h(t) = -\frac{1}{2} g t^{2} + V_{0} t + H_{0} \]Using feet, the gravity constant \( g \) is \( 32 \text{ ft/s}^2 \). The initial speed \( V_{0} = 64 \text{ ft/s} \), and the object touches the ground (\( h = 0 \)) at \( t = 5 \) seconds. \( H_{0} \) represents the initial height, which is the height of the building.
Substitute the known values into the equation:
\[ 0 = -16(5)^{2} + (64)(5) + H_{0} \] \[ 0 = -16(25) + 320 + H_{0} \] \[ 0 = -400 + 320 + H_{0} \]Solving for \( H_{0} \):
\[ H_{0} = 400 - 320 = 80 \text{ feet} \]Problem 4: Initial Velocity from Maximum Height
From the ground, an object is thrown upward with an initial speed \( V_0 \). Three seconds later, it reaches its maximum height. What is the initial velocity \( V_0 \)?
View Step-by-Step Solution
The height \( h \) of a projectile thrown upward from the ground (initial height \( H_0 = 0 \)) is given by:
\[ h(t) = -16t^2 + V_0 t \]The maximum height for a quadratic function \( at^2 + bt + c \) occurs at the vertex time:
\[ t = \frac{-b}{2a} = \frac{-V_0}{2(-16)} = \frac{V_0}{32} \]We are given that this time is exactly 3 seconds. Set the equation equal to 3:
\[ \frac{V_0}{32} = 3 \]Solving for \( V_0 \), we find the initial velocity:
\[ V_0 = 3 \times 32 = 96 \text{ feet/second} \]Verification: Maximum height \( h(3) = -16(9) + 96(3) = -144 + 288 = 144 \) ft. ✓
Summary of Formulas
| Quantity | Formula | Notes |
|---|---|---|
| Height of a projectile | \( h(t) = -\frac{1}{2} g t^2 + V_0 t + h_0 \) | Quadratic in \( t \) |
| Free fall distance | \( d(t) = \frac{1}{2} g t^2 \) | When \( V_0 = 0 \) |
| Time to maximum height | \( t = -\frac{b}{2a} \) | Vertex of the parabola |
| Maximum height | \( h_{\max} = h(t_{\text{vertex}}) \) | Substitute vertex time |
| Time to hit ground | Solve \( h(t) = 0 \) | Take positive root |
| Gravity (English) | \( g = 32 \text{ ft/s}^2 \) | Use with feet |
| Gravity (metric) | \( g = 9.8 \text{ m/s}^2 \) | Use with meters |
Frequently Asked Questions
What is the formula for the height of a projectile?
If air resistance is ignored, the height \( h(t) \) of a projectile is given by \( h(t) = -\frac{1}{2}gt^2 + V_0 t + h_0 \), where \( g \) is the acceleration due to gravity (\( 32 \text{ ft/s}^2 \) or \( 9.8 \text{ m/s}^2 \)), \( V_0 \) is the initial velocity, and \( h_0 \) is the initial height.
How do you find the maximum height of a projectile?
For a quadratic height function \( h(t) = at^2 + bt + c \), the maximum height occurs at the vertex of the parabola, which is reached at time \( t = -b/(2a) \). Substitute this time back into \( h(t) \) to find the maximum height.
How do you find when a projectile hits the ground?
Set \( h(t) = 0 \) and solve the resulting quadratic equation for \( t \). Discard any negative solution (time cannot be negative) and keep the positive root. This gives the time of flight until impact.
What is the acceleration due to gravity in projectile motion problems?
In the English system (feet), \( g \approx 32 \text{ ft/s}^2 \). In the metric system (meters), \( g \approx 9.8 \text{ m/s}^2 \). The choice of units determines which value to use; mixing units will give incorrect results.
How do you find the time interval during which a projectile is above a certain height?
Set \( h(t) \) greater than the target height and solve the resulting quadratic inequality. The two roots define the interval; the projectile is above the target height between the smaller and larger root.
How do you solve a well-depth problem using the speed of sound?
Let \( T_1 \) be the fall time and \( T_2 \) the sound travel time, so \( T_1 + T_2 = \) total time. The depth satisfies \( H = \frac{1}{2}gT_1^2 \) and \( H = c \cdot T_2 \) where \( c \) is the speed of sound. Substitute \( T_2 = T_{\text{total}} - T_1 \) and solve the resulting quadratic for \( T_1 \), then compute \( H \).