Question

Philip jumps up with to a height of 3 m above the ground. What was Philip's initial velocity? round to the tenth.

Answer 1

The initial velocity of Philip was 7.66 m/s

Given data:

The vertical height is h=3 m.

Considering ground as the reference, then the initial potential energy of Philip is zero, i.e., PEi=0

The formula for the kinetic energy is given by,

`KE_i=(1)/(2)mv^2`

Here, m is mass and v is the velocity.

After reaching the height of 3 m Philip comes to a stop. It means the final kinetic energy is zero, i.e. KEf=0.

The final potential energy is given by,

`PE_f=mgh`

Here, g is the gravitational acceleration.

Applying the conservation of energy between initial position and final position.

`\begin{gathered} KE_i+PE_i=KE_f+PE_f \\ (1)/(2)mv^2+0=0+mgh \\ v=\sqrt[]{2gh} \\ v=\sqrt[]{2*9.8*3} \\ v=7.66\text{ m/s} \end{gathered}`

Thus, the initial velocity of Philip was 7.66 m/s.