Example from Physics, 5.3 Projectile Motion

A Fireworks Projectile Explodes High and Away (part 1)

During a fireworks display like the one illustrated in Figure 5.30, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75° above the horizontal. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

The diagram shows the trajectory of a fireworks shell. An x-axis is labeled x, a y-axis is labeled h, and a vector, labeled V o, points upward from the axis origin. An angle is formed by the vector and the x-axis with a measure of seventy-five degrees.

Figure 5.30 The diagram shows the trajectory of a fireworks shell.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
The motion can be broken into horizontal and vertical motions in which $a_{x} = 0$ and $a_{y} = g$ . We can then define $x_{0}$ and $y_{0}$ to be zero and solve for the maximum height.
Solution
Solution for (a)

By height we mean the altitude or vertical position $y$ above the starting point. The highest point in any trajectory, the maximum height, is reached when $v_{y} = 0$ ; this is the moment when the vertical velocity switches from positive (upwards) to negative (downwards). Since we know the initial velocity, initial position, and the value of v_y when the firework reaches its maximum height, we use the following equation to find $y$
$v_{y}^{2} = v_{0 y}^{2} - 2 g (y - y_{0}) .$
Because $y_{0}$ and $v_{y}$ are both zero, the equation simplifies to
$0 = v_{0 y}^{2} - 2 g y .$
Solving for $y$ gives
$y = \frac{v_{0 y}^{2}}{2 g} .$
Now we must find $v_{0 y}$ , the component of the initial velocity in the y-direction. It is given by $v_{0 y} = v_{0} \sin θ$ , where $v_{0}$ is the initial velocity of 70.0 m/s, and $θ = 75^{\circ}$ is the initial angle. Thus,
$v_{0 y} = v_{0} \sin θ_{0} = (70.0 m/s) (\sin 75^{\circ}) = 67.6 m/s$

and $y$ is
$y = \frac{{(67.6 m/s)}^{2}}{2 (9.80 {m/s}^{2}),}$
so that

$y = 233 m .$

0 of 2 steps open

How did it go?

Strategy

Solution

Strategy

Solution