Launch angles that will allow a ball to clear a wall?

Question

A cannonball is shot with an initial speed of 50 m/s (at a launch angle of 40 degrees with the horizontal) toward a castle wall 220 m away. The height of the wall is 30 m. Ignore air resistance. What launch angles will allow the cannonball to just barely clear the top of the castle wall?

Thanks for any help! =)

Answer 1

There are two elements here:

1. In the vertical direction, we have motion under constant acceleration:

Vf = Vi + A T

D = Vi T + (1/2) A T^2

where:

T is the time interval

A is the acceleration ( = g, about 9.8 m/s^2)

Vi is the initial velocity

Vf is the final velocity

D is the distance traveled

2. In the horizontal direction, there is no acceleration so the velocity is constant so:

distance = speed x time

If @ is the angle of launch, then the initial speed of the projectile can be split into the horizontal and vertical components

Vh = V cos @

Vv = V sin @

The total time Tt is such that the the cannonball reaches the castle wall. That is:

220 = Vh Tt

In that time the cannonball rises to its peak elevation and then falls the same distance minus 30 m.

Tu, the time going up, can be computed in terms of Vv by using the velocity equation.

0 = Vv - g Tu

Then you can compute the height the cannonball reaches in terms of Vv by using the distance equation:

height = Vv - (1/2) g Tu^2

Then you can compute the time to fall to the castle wall with the distance equation:

height - 30 = 0 + (1/2) g Td^2

And of course, the total time is the time up plus the time down:

Tt = Tu + Td

The only other equation you need is the trigonometric identity for sin and cos:

sin^2 X + cos^2 X = 1, for any X

Solving these equations will give you sin @ and cos @. Then you can use the arctan function to determine @