Projectile Problems With Solutions - Free Mathematics Tutorials

Projectile Motion Problems with Step-by-Step Solutions for Students

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.

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

\[ h(t) = -\tfrac{1}{2} g t^{2} + V_{0} t + h_{0} \]

where \( g \) is the acceleration due to gravity (approximately \( 32 \,\text{ft/s}^{2} \) on Earth), \( V_{0} \) is the initial velocity at \( t = 0 \), and \( h_{0} \) is the initial height at \( t = 0 \).

For an object that is dropped (initial velocity zero), the distance traveled is

\[ h(t) = \tfrac{1}{2} g t^2 \]

For a quadratic function of the form

\[ h(t) = a t^2 + b t + c \]

with a negative coefficient \( a \), the maximum height occurs at \[ t = - \dfrac{b}{2 a} \] which corresponds to the vertex of the parabola representing \( h(t) \).

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 \( 80 \) feet building.

1. What is the highest point reached by the object?
2. How long does it take the object to reach its highest point?
3. After how many seconds does the object hit the ground?
4. For how many seconds is the height of the object higher than 90 feet?

Show Solution Solution:

1. 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 \( h \) occurs 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 \[ t = \frac{-b}{2a}. \] Hence, for \( h \) given above, the vertex is at \[ t = \frac{-32}{2(-16)} = 1 \ \text{second}. \] One 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 \ \text{feet}.
2. It takes \(1\) second for the object to reach its highest point.
3. At the ground, \( h = 0 \). Hence, the solution of the equation \( h = 0 \) gives the time \( t \) at which the object hits the ground: \[ -16t^{2} + 32t + 80 = 0. \] The above quadratic equation has two solutions: one negative and the other positive (approximately equal to \( 3.5 \) seconds). So, it takes \( 3.5 \) seconds for the object to hit the ground after it has been thrown upward. The graphical meanings of the answers to parts a, b, and c are shown below.
4. The object is higher than \( 90 \) feet for all values of \( t \) satisfying the inequality \( h > 90 \): \[ -16t^{2} + 32t + 80 > 90. \] The above inequality is satisfied for \[ 0.4 \lt t \lt 1.6 \quad \text{(seconds)}. \] The height of the object is higher than \( 90 \) feet for \[ 1.6 - 0.4 = 1.2 \ \text{seconds}. \] The graph below shows \( h \) in terms of \( t \) and clearly indicates that \( h > 90 \) for \( 0.4 \lt t \lt 1.6 \).

Problem 2 - Rock Dropped into a Well

A rock is dropped into a well and the distance traveled is \[ d = 16t^{2} \quad \text{(in feet),} \] where \(t\) is the time in seconds. If the water splash is heard \(3\) seconds after the rock was dropped, and the speed of sound is \(1100 \ \text{ft/sec},\) approximate the height of the well.

Show Solution Solution:

Let \(T_{1}\) be the time it takes the rock to reach the bottom of the well. If \(H\) is the height of the well, then \[ H = 16 T_{1}^{2}. \] Let \(T_{2}\) be the time it takes the sound wave to reach the top of the well. Then \[ H = 1100 T_{2}. \] The relationship between \(T_{1}\) and \(T_{2}\) is \[ T_{1} + T_{2} = 3. \] Eliminate \(H\) \[ 16 T_{1}^{2} = 1100 T_{2}. \] Substitute \(T_{2} = 3 - T_{1}\): \[ 16 T_{1}^{2} = 1100 (3 - T_{1}). \] Form the quadratic equation \[ 16 T_{1}^{2} + 1100 T_{1} - 3300 = 0. \] Solve for \(T_{1}\) The quadratic has two solutions, but only one is positive: \[ T_{1} \approx 2.88 \ \text{seconds}. \] Calculate the height of the well \[ H = 16 T_{1}^{2} = 16 (2.88)^{2} \approx 132.7 \ \text{feet}. \]

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 \(5 \, \text{seconds}\) later. What is the height of the building?

Show Solution Solution:

The formula for the height \( h \) of a projectile thrown upward is given by \[ h(t) = -\tfrac{1}{2} g t^{2} + V_{0} t + H_{0} \] where \( g \) is a constant equal to \( 32 \). The initial speed \( V_{0} \) is known, and when \( t = 5 \) seconds the projectile touches the ground, i.e., \( h = 0 \). Here, \( H_{0} \) is the initial height (the height of the building). Thus, we can write \[ 0 = -\tfrac{1}{2}(32)(5)^{2} + (64)(5) + H_{0} \] Solving for \( H_{0} \), we get the initial height which is the height of the building. \[ H_{0} = 80 \ \text{feet} \]

Problem 4 - Initial Velocity from Maximum Height

From ground, an object is thrown upward with an initial speed \( V_0 \). Three seconds later, it reaches a maximum height. What is the initial velocity \( V_0 \)?

Show Solution Solution:

The height \( h \) of a projectile thrown upward from the ground (initial height \( H_0 = 0 \)) is given by \[ h(t) = -\tfrac{1}{2}(32)t^2 + V_0 t \] which simplifies to \[ h(t) = -16t^2 + V_0 t. \] The maximum height occurs at \[ t = \frac{-V_0}{2(-16)}. \] But this time is given to be \( 3 \) seconds. Hence \[ \frac{-V_0}{2(-16)} = 3. \] Solving for \( V_0 \), we obtain the initial velocity \[ V_0 = 96 \ \text{feet/second}. \]