Question

Josh fires a catapult 18 ft above the ground with an initial velocity of 25 ft/sec. The misslie leaves the catapult at an angle of degree theta with the hoizontal and heads toward a 40 ft wall 500 ft from the launch site. If the missle is fired at a 20 degree angle, does it clear the wall? Justify the answer

If the missle is fired at a 35 degree angle, how high does it go?

Answer 1

The vertical component of the velocity is 25sin20° = 8.55

So, the height y is
y = 18 + 8.55t - 16t^2
If the missile just clears the wall, then
40 = 18 + 8.55t - 16t^2
Hmm. No solutions.

Check for typos, and take a look at wikipedia's article on "trajectory" for details of the range, height, etc.

Answer 2

initial horizontal velocity: 25*cos(theta)=v1
initial vertical velocity:
25*sin(theta)=v2

where theta is the angle to the horizontal.

max height:
(v2)^2-2*32ft/s^2*(height)=0

will it go over the wall:
v1*t=500 ft

solve the above equation for t and plug into:
18+v2*t-(32ft/s^2)/2*t^2= height at given time

Answer 3

one correction:

"max height:
(v2)^2-2*32ft/s^2*(height)=0" is wrong

it should be:
(v2)^2-2*32ft/s^2*(height -18)=0