# Answer to Question #4570 in Mechanics | Relativity for Theresa

Question #4570

if you give a golf ball an intial speed of 130 mph at an angle of 60 degrees,

a. how far will the ball travel assuming level ground?

b. how long will it take the ball to come back down to the ground?

c. what is the peak height of the ball?

d. how far will the ball travel if it is hit from a 20 m cliff?

a. how far will the ball travel assuming level ground?

b. how long will it take the ball to come back down to the ground?

c. what is the peak height of the ball?

d. how far will the ball travel if it is hit from a 20 m cliff?

Expert's answer

if you give a golf ball an initial speed of V = 130 [miles per hour] = 58,1152 [meters per second] at an angle of α = 60 degrees,

a. how far will the ball travel assuming level ground?

Horizontal component of velocity Vh = V*cosα = 58,1152*cos60 = 58,1152/2 = 29,0576 m/s.

Vertical component of velocity Vv = V*sivα = 58,1152*sin60 = 58,1152*sqrt(3)/2 = 50,3292 m/s.

At the top point Vv-g*T = 0, where T is the time reached to get the peak height, so T = Vv/g = 50,3292/9.8 = 5,1356 s.

Peak height H = VvT - (gT^2)/2 = 50,3292*5,1356 - (9.8/2)*5,1356^2 = 129,2361 m.

Total time of flight is Tt = 2T = 2*5,1356 = 10,2712 s.

So, the distance is Tt*Vh = 10,2712*29,0576 = 298,4564 m.

b. how long will it take the ball to come back down to the ground?

As we already admitted, total time of flight Tt = 10,2712 s.

c. what is the peak height of the ball?

As we already admitted, peak height H = 129,2361 m.

d. how far will the ball travel if it is hit from a 20 m cliff?

We can submit that the ball has additional time to fall Ha = 20 m down. We can find this time Ta from equation

Vv*Ta+g*Ta^2 = Ha,

or

50,3292*Ta+9.8*Ta^2 = 20.

Solving this quadratic equation we obtain

Ta = 1,4825 s.

So, additional distance is

Da = Vv*Ta = 29,0576*1,4825 = 43,0779 m

and the total distance

Dt = Da+D = 43,0779+298,4564 = 341,5343 m.

a. how far will the ball travel assuming level ground?

Horizontal component of velocity Vh = V*cosα = 58,1152*cos60 = 58,1152/2 = 29,0576 m/s.

Vertical component of velocity Vv = V*sivα = 58,1152*sin60 = 58,1152*sqrt(3)/2 = 50,3292 m/s.

At the top point Vv-g*T = 0, where T is the time reached to get the peak height, so T = Vv/g = 50,3292/9.8 = 5,1356 s.

Peak height H = VvT - (gT^2)/2 = 50,3292*5,1356 - (9.8/2)*5,1356^2 = 129,2361 m.

Total time of flight is Tt = 2T = 2*5,1356 = 10,2712 s.

So, the distance is Tt*Vh = 10,2712*29,0576 = 298,4564 m.

b. how long will it take the ball to come back down to the ground?

As we already admitted, total time of flight Tt = 10,2712 s.

c. what is the peak height of the ball?

As we already admitted, peak height H = 129,2361 m.

d. how far will the ball travel if it is hit from a 20 m cliff?

We can submit that the ball has additional time to fall Ha = 20 m down. We can find this time Ta from equation

Vv*Ta+g*Ta^2 = Ha,

or

50,3292*Ta+9.8*Ta^2 = 20.

Solving this quadratic equation we obtain

Ta = 1,4825 s.

So, additional distance is

Da = Vv*Ta = 29,0576*1,4825 = 43,0779 m

and the total distance

Dt = Da+D = 43,0779+298,4564 = 341,5343 m.

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