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aliya0001 [1]
3 years ago
13

On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 2.50 times as far as he wo

uld have on earth, given the same initial velocities on both planets. The ball is launched at a speed of 44.4 m/s at an angle of 26 ° above the horizontal. When the ball lands, it is at the same level as the tee. On the distant planet, what are (a) the maximum height and (b) the range of the ball?

Physics
1 answer:
myrzilka [38]3 years ago
6 0

Answer:

a) On the distant planet, the ball will travel 395 m horizontally.

b) On the distant planet, the max-height will be 48.3 m.

Explanation:

The equation for the position of an object moving in a parabolic trajectory is the following:

r = (x0 + v0 • t • cos θ, y0 + v0 • t • sin θ + 1/2 • g • t²)

Where:

r = vector position at time t

x0 = initial horizontal position

v0 = initial velocity

t = time

θ = launching angle

y0 = initial vertical position

g = acceleration due to gravity

a) First, let´s calculate the horizontal distance traveled as if we would be on earth. For this, we have to find the magnitude of the vector "r" in the figure. Seeing the figure, we know that the y-component of the vector "r" is 0 if we place the center of the frame of reference at the launching point.

Then, using the equation for the y-component of "r":

y = y0 + v0 • t • sin θ + 1/2 • g • t² (y0 = 0)

0 m = 44.4 m/s · t · sin 26° - 1/2 · 9.8 m/s² · t²

4.9 m/s² · t² = 44.4 m/s · t · sin 26°

t = 44.4 m/s · sin 26° / 4.9 m/s²

t = 3.97 s

Now, we can calculate the x-component of the vector position at final time:

x = x0 + v0 • t • cos θ (x0 = 0)

x = 44.4 m/s · 3.97 s · cos 26° = 158 m

On the distant planet, the ball will travel 158 m · 2.50 = 395 m horizontally.

b) Since the trajectory is parabolic, the maximum height will be reached at the half of the flight. Then, the maximum height will be at t = 3.97 s/2 = 1.99 s.

The max-height on earth wuould be:

y = y0 + v0 • t • sin θ + 1/2 • g • t²  (y0 = 0)

y = 44.4 m/s · 1.99 s · sin 26° - 1/2 · 9.8 m/s² · (1.99 s)²

y = 19.3 m

On the distant planet the max-height will be 19.3 m · 2.50 = 48.3 m

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