Question

A cannonball is catapulted toward a castle. The cannonball's velocity when it leaves the catapult is 40 m/s at an angle of 37° with respect to the horizontal and the cannonball is 7.0 m above the ground at this time. (a) What is the maximum height above the ground reached by the cannonball? (b) Assuming the cannonball makes it over the castle walls and lands back down on the ground, at what horizontal distance from its release point will it land? (c) What are the x- and y-components of the cannonball's velocity just before it lands? The y-axis points up.

Answer 1

Answer:

a) Maximum height = 36.6 m

b) Horizontal distance at which the ball lands = 166.1 m

c) x-component = 32 m/s. y-component = - 27 m/s  

Explanation:

Please, see the attached figure for a description of the problem.

The velocity vector "v" of the cannonball has two components, a horizontal component, "vx", and a vertical component "vy". Notice that at the maximum height, the vertical component "vy" of the velocity vector is 0.

In the same way, the position vector "r" is composed by "rx", its horizontal component, and "ry", the vertical component.

The velocity vector "v" ad the position vector "r" at time "t" are given by the following equations:

v = (v0 * cos α, v0 * sin α + g * t)

r = (x0 + v0 * t * cos α, y0 + v0 * t * sin α + 1/2 * g * t²)

Where

v0 = magnitude of the initial velocity vector

α = launching angle

g = gravity acceleration (-9.8 m/s², because the y-axis points up)

t = time

x0 = initial horizontal position

y0 = initial vertical position

If we consider the origin of the system of reference as the point at which the cannonball leaves tha catapult, then, x0 and y0 = 0

a) We know that at maximum height, the vertical component of the vector "v" is 0, because the ball does not move up nor down at that moment (see figure). Then:

0 = v0 * sin α + g * t

-v0 * sin α / g = t

-40 m/s * sin 37° / -9.8 m/s² = t

t = 2.5 s

We can now calculate the position of the cannonball at time t=2.5 s to obtain the maximum height:

r = (x0 + v0 * t cos α, y0 + v0 * t * sin α + 1/2 * g * t²)

The max height is the magnitude of the vector ry max (see figure). The vector ry max is:

ry = (0, y0 + v0 t sin α + 1/2 g * t²)

magnitude of ry = $|ry|= \sqrt{(0m)^{2} + (y0 + v0* t*sin \alpha+ 1/2*g*t^{2})^{2}}= y0 + v0*t*sin \alpha + 1/2*g*t^{2}$)

Then:

max height = y0 + v0 * t * sin α + 1/2 * g * t²

max height = 0 m + 40 m/s * 2.5 s * sin 37° - 1/2* 9.8 m/s² * (2.5 s)² = 29.6 m

Since the ball leaves the catapult 7 m above the ground, the max height above the ground will be 29.6 m + 7 m = 36.6m

max height = 36.6 m

b) When the ball hits the ground, the position is given by the vector "r final" (see figure). The magnitude of "rx", the horizontal component of "r final", is the horizontal distance between the catapult and the wall.

r final = ( x0 + v0 * t * cos α, y0 + v0 * t * sin α + 1/2 * g * t²)

We know that the vertical component of "r final" is -7 (see figure).

Then, we can obtain the time when the the ball hits the ground:

y0 + v0 * t * sin α + 1/2 * g * t² = -7 m

0 m + 40 m/s * t * sin 37° + 1/2 g * t² = -7 m

7 m + 40 m/s * t * sin 37° + 1/2 (-9.8 m/s²) * t² = 0

7 m + 24.1 m/s * t - 4.9 m/s² * t² = 0

solving the quadratic equation:

t = 5.2 s (The negative solution is discarded).

With this time, we can calculate the value of the horizontal component of "r final"

Distance to the wall = |rx| = x0 + v0 t cos α

|rx| = 0m + 40 m/s * 5.2 s * cos 37° = 166.1 m

c) With the final time obtained in b) we can calculate the velocity of the ball:

v = (v0 * cos α, v0 * sin α + g * t)

v =(40 m/s * cos 37°, 40 m/s * sin 37°  -9.8 m/s² * 5.2 s)

v =(32 m/s, -27 m)

x-component = 32 m/s

y-component = - 27 m/s