You are trying to throw a ball from ground under a known angle theta 0 and known

Question

You are trying to throw a ball from ground under a known angle theta0 and known initial velocity v0 to a nearby window. The known gravitational acceleration constant is g. The height of the window is half the height of the maximum height of the trajectory with respect to the original thrown location.

show that the time tF to hit the window is given by: tF= ((v0 sin(theta))/g)(1+ sqrt(2)/2)

show that the distance xF from the thrown location in x where the ball hits the window is given by: ((v02 sin2(theta))/2g)(1+ sqrt(2)/2)

Explanation / Answer

max height = U^2sin^2(theta)/2g

new height = 1/2of max height = U^2sin^2(theta)/4g

u^2sin^2(theta)/4g = usin(theta)*t - 1/2gt^2

OR

1/2gt^2 -usin(theta)*t + U^2sin^2(theta)/4g = 0

solving above quadratic equation we get t = (usin(theta)/g)(1+sqrt2/2)

b)

X = ucos(theta)*t =    ucos(theta)*(usin(theta)/g)(1+sqrt2/2)

=> X = ((u2sin2(theta))/2g)(1+ sqrt(2)/2)

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