2 dimensional physics problem?

Question

A 1.2 kg golf club approaches a 0.047 kg golf ball that is resting on the ground. The ball remains in contact with the golf club for 0.050 s and during this time the ball experiences an acceleration of 520 m/s2. The ball is launched at an angle of 38° above the ground. Assume that the golf ball loses contact with the club at ground level (0 meter) and that it lands (and immediately stops) in a sand trap that is also at ground level.

-How much time passes before the ball reaches its peak height?

-What is the ball's total time of flight?

-What is the ball's horizontal displacement between the contact point and the ball's landing spot?

-What is the ball's speed at its maximum height?

Is the 520m/s^2 horizontal acceleration or the acceleration at that angle? I'm really not sure how to start this problem, so any help would be appreciated.

Answer 1

the purpose of the acceleration information is to allow you to calculate the initial speed of the ball

if the ball accelerated at the rate of 520m/s/s for 0.05 s, it achieved a speed of 520x0.05=26 m/s

the ball left the ground at an angle of 38 degrees with this speed, so its horizontal speed is 26cos38

the initial vertical speed is 26sin38

the ball will reach max height when its vertical speed is zero; the vertical speed decreases at the rate of 9.8m/s/s, so the ball is at apex when t=26sin38/9.8 = 1.63 s

with no friction, the time to fall from this height to the ground is the same, and so the total time in the air is 3.26 s

with no friction, the ball's horizontal speed is the same (since there are no forces in the horizontal direction), so the horizontal distance traveled is just the horizontal speed x total time

or 26cos38x3.26 s = 66.8m

at max height, there is no vert velocity, so the only velocity is horizonatl which is 26cos38

Answer 2

I think that the acceleration is at that angle, 38° above the ground. If it were horizontal, the ball would be launched in horizontal direction.