Question

Olympic diver Matthew Mitcham springs upward from a diving board that is 3.40 m above the water. He enters the water at a 80.0 degree angle with respect to the water surface, at a speed of 9.32 m/s.
a)Determine the magnitude of his initial velocity.

b)Determine the direction of his initial velocity, in terms of degrees relative to horizontal.

c)Determine his maximum height above the water.

Answer 1

Explanation:

Given that,

Height = 3.40 m

Speed of water = 9.32 m/s

Angle = 80.0°

Suppose the angle at diving board is 10°.

(a). We need to calculate the magnitude of his initial velocity

Using the formula of horizontal velocity

$u\cos\theta=u\cos\theta$

$u=\dfrac{u\cos\theta}{u}$

Put the value into the formula

$u=\dfrac{9.32\cos80}{\cos10}$

$u=1.64\ m/s$

(b). We need to calculate the direction of his initial velocity,

The direction of his initial velocity is 10° east of south.

(c). We need to calculate the time

Using formula of time

$t=\dfrac{u\sin\theta}{g}$

Put the value into the formula

$t=\dfrac{1.64\sin10}{9.8}$

$t=0.029\ sec$

We need to calculate his maximum height above the water

Using equation of motion

$S_{y}=u_{y}+\dfrac{1}{2}gt^2$

Put the value into the formula

$S_{y}=1.64\sin10+\dfrac{1}{2}\times9.8\times(0.029)^2$

$S_{y}=0.288\ m$

Hence, This is the required solution.