
Review Projectile Equations
If air resistance is ignored, the height h of a projectile above the ground after t seconds is given by
H(t) =  (1 / 2) g t^{ 2} + Vo t + Ho
where g is the acceleration due to gravity which on earth is approximately equal to 32 feet / s^{ 2}, Vo is the initial velocity (when t = 0 ) and Ho is the initial height (when t = 0).
If an object is dropped, the distance traveled by this object is given by
H(t) = (1 / 2) g t^{ 2}
For a quadratic function of the form
H(t) = a t^{ 2} + b t + c
and with a negative, the maximum of H occur at t =  b / 2a (vertex of the graph of H).
Projectile Problems with Detailed Solutions
Problem 1:
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 feet building.
a  What is the highest point reached by the object?
b  How long does it take the object to reach its highest point?
c  After how many seconds does the object hit the ground?
d  For how many seconds is the hight of the object higher than 90 feet?
Solution to Problem 1:
 a  The height h given above is a quadratic function. The graph of h as a function of time t gives a parabolic shape and the maximum height h occur at the vertex of the parabola. For a quadratic function of the form h = a t^{ 2} + b t + c, the vertex is located at t =  b / 2a. Hence for h given above the vertex is at t
t = 32 / 2(16) = 1 second.
 1 second after the object was thrown, it reaches its highest point (maximum value of h) which is given by
h = 16 (1)^{ 2} + 32 (1) + 80 = 96 feet
 b  It takes 1 second for the object to reach it highest point.
 c  At the ground h = 0, hence the solution of the equation h = 0 gives the time t at which the object hits the ground.
16 t^{ 2} + 32 t + 80 = 0
 The above quadratic equation has two solutions one is negative and the second one is positive and approximately equal to 3.5 seconds. So it takes 3.5 seconds fopr the object to hit the ground after it has been thrown upward. The graphical meanings to the answers to parts a, b and c are shown below.
.
 d  The object is higher than 90 feet for all values of t statisfying the inequality h > 90. Hence
16 t^{ 2} + 32 t + 80 > 90
 The above inequality is satisfied for
0.4 < t < 1.6 (seconds)
 The height of the object is higher than 90 feet for
1.6  0.4 = 1.2 seconds.
 The graph below is that of h in terms of t and clearly shows that h is greater than 90 feet for t between 0.4 and 1.6 seconds.
.
Problem 2:
A rock is dropped into a well and the distance traveled is 16 t ^{ 2} feet, where t is the time. If the water splash is heard 3 seconds after the rock was dropped, and that the speed of sound is 1100 ft / sec, approximate the height of the well.
Solution to Problem 2:
 Let T1 be the time it takes the rock to reach the bottom of the well. If H is the height of the well, we can write
H = 16 T1^{ 2}
 Let T2 be the time it takes sound wave to reach the top of the well. we can write
H = 1100 T2
 The relationship between T1 and T2 is
T1 + T2 = 3
 Eliminate H and combine the equations H = 16 T1^{ 2} and H = 1100 T2 to obtain
16 T1^{ 2} = 1100 T2
 We now substitute T2 by 3  T1 in the above equation
16 T1^{ 2} = 1100 (3  T1)
 The above is a quadratic equation that may be written as
16 T1^{ 2} + 1100 T1  3300 = 0
 The above equations has two solutions and only one of them is positive and is given by
T1 = 2.88 seconds (2 decimal places)
 We now calculate the height H of the well
H = 16 T1^{ 2} = 16 (2.88)^{ 2} = 132.7 feet
Problem 3:
From the top of a building, an object is thrown upward with an initial speed of 64 ft/sec. It touches the ground 5 seconds later. What is the height of the bulding
Solution to Problem 3:
 The formula for the height H of a projectile thrown upward is given by
H(t) = (1 / 2) g t^{ 2} + Vo t + Ho
 g is a constant and equal to 32. The initial speed Vo is known and also when t = 5 seconds H = 0 (touches the ground). Ho is the initial height or height of the building. Hence we can write
0 = (1 / 2) (32) (5)^{ 2} + (64) (5) + Ho
 Solve the above for Ho to find the height of the building
Ho = 80 feet
Problem 4:
From ground, an object is thrown upward with an initial speed Vo. Three seconds later, it reaches a maximum height. What is the initial velocity Vo?
Solution to Problem 4:
 The height H of a projectile thrown upward from ground (initial height = 0 or Ho = 0) is given by
H(t) = (1 / 2) (32) t^{ 2} + Vo t
H(t) = 16 t^{ 2} + Vo t
 The maximum height occurs at t = Vo / 2(16). But this time is given and is equal to 3 seconds. Hence
Vo / 2(16) = 3
 Solve the above to find
Vo = 96 feet / sec
More References and Links
Projectile Problems with Solutions and Explanations
Solutions and Explanations to Projectile Problems
Projectile Motion Calculator and Solver
math problems with detailed solutions .
