Jessica attains a height of 4.7 feet above the launch and landing ramps after 1 second. Her initial velocity is 25 feet per second. Find the angle of her launch.

Question

a. Which equation can you use with the given information to solve for theta ? b. Substitute the known values and solve for theta.
What is Jessica’s height above the launch and landing ramps after 0.5 second?
What distance has Jessica traveled after 1 second?

Answer 1

I am assuming the landing ramp is at the same height as the launch ramp. So, we can work as though that's y=0

y(t) = (v sinθ)t - 16t^2
So, with y(1) = 4.7,
25sinθ - 16 = 4.7

Now you can find y for any t.

The horizontal speed is a constant 25 cosθ
so, x(t) = (25cosθ)t

Answer 2

Vo = 25ft./s[Ao].
Xo = 25*cosA.
Yo = 25*sinA.

Y^2 = Yo^2 + 2g*h = 0.
Yo^2 +(- 64)*4.7 = 0,
Yo = 17.3 ft/s. = vert. component of initial velocity.
a. Yo = 25*sinA = 17.3 ft/s.
b. A = 43.8o = angle of her launch.

h = Yo*t + 0.5g*t^2 = 17.3*0.5 + (-16)*0.5^2 = 4.65 ft.

d = Xo * t = 25*cos43.8 * 1 =