Two friends are playing golf. The first friend hits a golf ball on level ground

Question

Two friends are playing golf. The first friend hits a golf ball on level ground with an initial speed of 35.0 m/s at an angle of 36.0° above the horizontal. (a) Assuming that the ball lands at the same height from which it was hit, how far away from the golfer does it land? Ignore air resistance. (m) (b) The second friend hits his golf ball with the same initial speed as the first, but the initial velocity of the ball makes an angle with horizontal that is greater than 45.0°. The second ball, however, travels the same horizontal distance as the first, and it too lands at the same height from which it was hit. What was the angle above horizontal of the initial velocity of this second golf ball? Ignore air resistance. (° above the horizontal)

Explanation / Answer

Range or horizontal distance=u2sin2 theta/g

For first golf ball R=35×35sin2×36/9.8=118.9m

For second golf ball R=118.9

35×35sin2theta/9.8=118.9

Theta=54

For two complementry angles range wil be same u2sin 2(90-theta)/g=u^2sin(180-2theta)/g=u^2sin 2 theta/g

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