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

Step-by-step 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=(u\cos\theta)/(u)`

Put the value into the formula

`u=(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=(u\sin\theta)/(g)`

Put the value into the formula

`t=(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)+(1)/(2)gt^2`

Put the value into the formula

`S_(y)=1.64\sin10+(1)/(2)*9.8*(0.029)^2`

`S_(y)=0.288\ m`

Hence, This is the required solution.