A seasoned mini golfer is trying to make par on a tricky hole #5. The golfer can

Question

A seasoned mini golfer is trying to make par on a tricky hole #5. The golfer can complete the hole by hitting the ball from the flat section it lays on, up a 45° ramp launching the ball into the hole which is d = 1.90 m away from the end of the ramp. If the opening of the hole and the end of the ramp are at the same height, y = 0.840 m, at what speed must the golfer hit the ball to land the ball in the hole? Assume a frictionless surface (so the ball slides without rotating). The acceleration due to gravity is 9.81 m/s2.

Explanation / Answer

Here ,

let the height at the top of incline is v

initial speed of the ball is u

for distance , d = 1.90 m

the range of ball is given as

R = v^2 * sin(2 * theta)/g

1.90 = v^2 * sin(2 * 45)/9.8

v = 4.32 m/s

Now , Usign third equation of motion for the initial velocity of ball

v^2 - u^2 = 2 *a * h

4.32^2 - u^2 = -2 * 9.8 * 0.84

u = 5.92 m/s

the initial speed of the ball is 5.92 m/s

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