Answer to Question #188463 in Mechanics | Relativity for Rhaesha Anjela Yu

Question

1. If Andrew Wiggins highest vertical leap is 1.29 m, then what is his takeoff speed and his hang time (total time to move upwards to the peak and then return to the ground)?
2. A canon ball leaves a cannon with velocity of 521 m/s. While accelerating, the canon ball moves a distance of 0.840 m. Determine the acceleration of the canon ball (assume a uniform acceleration).

Answer 1

1.

Let v denotes the take off velocity required for jump.

 T denotes the required time.

M denotes the mass of Andrew.

Using energy conservation-

$0+\dfrac{1}{2}(Mv^2)=M\times g\times 1.29+0$

So we get, $v^2=2\times 1.29\times 9.8$

       $v=\sqrt{25.31}=5.03m/s$

Time taken t move upwards-

using first equation of motion-

$v=u-gt 0=5.03-9.8t\Rightarrow t=\dfrac{5.03}{9.8}=0.51 s$

Total hang time $= 2t=2(0.51)=1.02 s$

2. As accelaration is constant, average velocity $v_{avg}$ is exactly half of peak velocity.

 so, $v_{avg}=\dfrac{521}{2}=260.5m/s$

 Time $\Delta t$ to cover distance d is therefore-

  $\Delta t=\dfrac{d}{v_{avg}}=\dfrac{0.840}{260.5}=0.003225$ seconds

So Accelaration becomes-

$a=\dfrac{\Delta v}{\Delta t}=\dfrac{521}{0.003225}=161,570 m/s^2$