Question

A diver springs upward from a board that is 3.90 m above the water. At the instant she contacts the water her speed is 13.2 m/s and her body makes an angle of 81.8 ° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.

Answer 1

a) u = 9.88 m/s

b)
`\theta = 79^0`

Step-by-step explanation:

given,

speed of the diver = 13.2 m/s

Angle made w.r.t horizontal = 81.8 °

The horizontal component of final velocity

v_h = 13.2 cos 81.8 °

The vertical component of final velocity

v_v = 13.2 sin 81.8 °

using equation of motion

initial velocity of vertical component

`u_v = √(v_v^2-2as)`

`u_v = √((13.2 sin 81.8^0)^2-2(-9.8)(-3.9))`

= 9.70 m/s

the horizontal component of the initial velocity

u_h = v_h = 13.2 cos 81.8 ° = 1.88 m/s

magnitude

`u = √(9.7^2+1.88^2)`

u = 9.88 m/s

direction

`\theta = tan{-1}((9.7)/(1.88))`

`\theta = 79^0`