Posted by bob on Sunday, January 20, 2013 at 5:48pm.
y = y0 + vy0*t -1/2*g*t^2
x = x0 + vx0*t
where y is the y position as a function of time, y0 is the initial y position, vy0 is the initial y velocity, g is the acceleration due to gravity (9.8 m/s^2), t is time, x is the x position as a function of time, vx0 is the initial x velocity. Plugging in the numbers for when the ball clears the wall:
20.6 = 1.1 + v0*sin(35)*t - 1/2*9.8*t^2
132 = 0 + v0*cos(35)*t
Use algebra to solve this system of 2 equations with 2 unkowns: the time t, and the velocity v0.
The components of the velocity (vy = component, vx = x component) at this time are given by:
vy = dy/dt = v0y - g*t = v0*sin(35) - g*t
vx = dx/dt = v0x = v0*cos(35)
Solve for vy and vx at the time you found in part b
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