Question

Two friends are playing golf. The first friend hits a golf ball on level ground with an initial speed of 47.0 ms at an angle of 35.0∘ 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.
(b) The second friend hits another 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 degrees. The second ball, however, travels the same horizontal distance as what you found in part (a), and it too lands at the same level from which it was hit. What was the angle in degrees above horizontal of the initial velocity of this second golf ball? Ignore air resistance.

Answer 1

A) 211.81 m

B) 65°

Step-by-step explanation:

We are told that the first friend hits a golf ball on level ground with an initial speed of 47 m/s at an angle of 35° above the horizontal.

Thus;

speed; u = 47 m/s

Projection angle; θ = 35°

A) We want to find how far away from the golfer the ball lands. This is simply the range. Range in projectile motion is given by the formula;

R = (u²•sin 2θ)/g

Plugging in the relevant values, we have;

R = (47² × sin(2 × 35))/9.8

R = 211.81 m

B) We are told that the second ball, travels the same horizontal distance as what is found in part (a).

Thus, R remains 211.81 m

Thus, the angle above horizontal of the initial velocity of this second golf ball will be;

90° - projection angle(θ) = 90 - 35 = 65°