Question

Interactive LearningWare 3.2 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 3.90 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 48.3 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?

Answer 1

Step-by-step explanation:

Given

Initial velocity(u)=48.3 m/s

Launch angle
`=26^(\circ)`

On earth Range
`=(u^2sin^2\theta )/(2g)`---1

g on distant planet is different thus

`R'=(u^2sin^2\theta )/(2g')`----2

Divide 2 & 1

`(R')/(R)=(g)/(g')`

`3=(g)/(g')`

`g'=(g)/(3)`

(a)maximum height

`h_(max)=(u^2sin^2\theta )/(2(g)/(3))`

`h_(max)=(48.3^2* \sin^226* 3)/(2* 9.8)`

`h_(max)=156.53 m`

(b)Range of ball

`R=(u^2sin2\theta )/((g)/(3))`

`R=(48.3^2* \sin52* 3)/(9.8)`

`R=562.18 m`