Question

provides a review of the concepts that are important in this problem. On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 4.80 times as far as he would have on earth, given the same initial velocities on both planets. The ball is launched at a speed of 44.9 m/s at an angle of 35 ° 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?

Answer 1

a) y = 162.6 m , b) R = 928.64 m

Step-by-step explanation:

We will solve this exercise using projectile launch kinematics, as the initial velocities and angle are equal on the planet and the Earth, let's look for the gravedd coinage of the planet

Earth

R = v₀² sin 2θ / g

R g = v₀² sin 2θ

In the planet

4.8 R = v₀² sin 2θ /
`g_(p)`

4.8 R
`g_(p)` = v₀² sin 2θ

4.8 R
`g_(p)` = R g

`g_(p)` = g / 4.8

`g_(p)` = 9.8 / 4.8

`g_(p)` = 2.04 m / s²

Now we can answer the questions

a) The maximum height

Vy² =
`v_(oy)`² - 2 g y

For ymax the vertical speed is zero (
`v_(y)` = 0)

sin θ =
`v_(oy)`/
`v_(o)`

`v_(oy)` =
`v_(o)` sinθ

`v_(oy)` = 44.9 sin 35

`v_(oy)` = 25.75 m/s

y =
`v_(oy)` ²/2
`g_(p)`

y = 25.75² / (2 2.04)

y = 162.6 m

b) the scope

R = v₀² sin 2θ /
`g_(p)`

R = 44.9² sin 2 35 / 2.04

R = 928.64 m