Question

remarks
although modeling the long - jumpers motion as that of a projectile is an oversimplification, the values obtained are reasonable.
question
how would the time of the jump and the horizontal distance traveled change if g were changed, for example if the jump could be repeated with the same initial velocity on a different planet? (select all that apply.)
the time of the jump increases when g is smaller.
increasing the time of the jump has no effect on the displacement.
the time of the jump decreases when g is smaller.
the displacement increases with increased time of the jump.
the displacement decreases with increased time of the jump.
practice it
use the worked example above to help you solve this problem. a long - jumper (as shown in the figure) leaves the ground at an angle of 17.0° to the horizontal and at a speed of 10.8 m/s.
(a) how long does it take for her to reach maximum height?
0.320 s
(b) what is the maximum height?

Explanation:

Step1: Analyze vertical - motion formula for time

The time \(t\) to reach maximum height in vertical - motion under gravity is given by \(t=\frac{v_{0y}}{g}\), where \(v_{0y} = v_0\sin\theta\) (initial vertical velocity) and \(g\) is the acceleration due to gravity. When \(g\) is smaller, with \(v_{0y}\) constant, the value of \(t\) (time of jump to reach maximum height) increases. So, the time of the jump increases when \(g\) is smaller.

Step2: Analyze horizontal - displacement formula

The horizontal displacement \(x = v_{0x}t\), where \(v_{0x}=v_0\cos\theta\) (initial horizontal velocity) and \(t\) is the time of flight. Since \(x\) is directly proportional to \(t\) (with \(v_{0x}\) constant), the displacement increases with increased time of the jump.

Answer:

The time of the jump increases when \(g\) is smaller.
The displacement increases with increased time of the jump.