Answer to Question #231391 in Physics for Moga

Question

Samuel leaves the end of a 4.0-m-high diving board and strikes the 

water 1.3 s later, 3.0 m beyond the end of the board. Considering the 

diver as a particle, determine: 

(a) His initial velocity,

 (b) The maximum height reached;

 (c) The velocity Vf

 with which he enters the water.

Answer 1

Given:

$H=4.0\:\rm m$

$L=3.0\:\rm m$

$\tau=1.3\:\rm s$

The equations of motion of a diver

$$x(t)=v_0\cos\theta *t\\ y(t)=H+v_0\sin\theta*t-\frac{gt^2}{2}$$

We get:

$$L=v_0\cos\theta *\tau\\ 0=H+v_0\sin\theta*\tau-\frac{g\tau^2}{2}$$

$$3=v_0\cos\theta *1.3\\ 0=4+v_0\sin\theta*1.3-\frac{9.8*1.3^2}{2}$$

$$v_0\cos\theta=2.3\\ v_0\sin\theta=3.3$$

(a)

$$v_0=\sqrt{2.3^2+3.3^2}=4.0\:\rm m/s$$ $$\sin\theta=3.3/4.0=0.82$$

(b)

$$v_0\sin\theta-gt^*=0\\ t^*=3.3/9.8=0.34\:\rm s$$

$$H_{\rm max}=y(t^*)=H+v_0\sin\theta*t^*-\frac{gt^{*2}}{2}\\ =4+3.3*0.34-\frac{9.8*0.34^2}{2}=4.6\:\rm m$$

(c)

$$v_f=\sqrt{v_0^2+2gH}=\sqrt{4.0^2+2*9.8*4}=9.7\:\rm m/s$$