A golfer hits a tee shot off of a cliff that is 18.2 m high. If it lands in the

Question

A golfer hits a tee shot off of a cliff that is 18.2 m high.  If it lands in the fairway a horizontal distance of 270 m from the tee and is in the air for 5.3 seconds,

a. What was the vertical component of the initial velocity of the golf ball?

b. What was the horizontal component of the velocity of the golf ball?

c. What was the initial velocity of the golf ball?

A golfer hits a tee shot off of a cliff that is 18.2 m high. If it lands in the fairway a horizontal distance of 270 m from the tee and is in the air for 5.3 seconds, What was the vertical component of the initial velocity of the golf ball? What was the horizontal component of the velocity of the golf ball? What was the initial velocity of the golf ball?

Explanation / Answer

a)

for vertical direction,

s=ut+(0.5)at^2

=>-18.2=Vy(5.3)-4.9*(5.3)^2

=>Uy=22.54 m/s

Vy=22.54-9.8t

b)

horizontal component of velocity,Vx

=>Vx*5.3=270

=>Vx=50.94 m/s

c)

initial velocity= (50.94 i+22.54 j) m/s

speed initial=55.70 m/s

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