Question

A ball is thrown straight up with an initial speed of 16.9 m/s. At what height above its initial position will the ball have one‑half its initial speed?

Answer 1

To find the height where the ball has one-half its initial speed, we can use the equations vf = v0 + gt and d = v0t - 0.5gt2.

Step-by-step explanation:

To find the height above its initial position where the ball has one-half its initial speed, we need to use the fact that the initial velocity (v0) of the ball is 16.9 m/s. At the highest point of the ball's trajectory, the velocity will be zero. We can use the formula vf = v0 + gt, where vf is the final velocity, g is the acceleration due to gravity, and t is the time it takes for the ball to reach its highest point.

By substituting vf = 0 and v0 = 16.9 m/s, we can solve for t. Once we have the value of t, we can use the equation d = v0t - 0.5gt2 to calculate the height (d) above the initial position where the ball will have one-half its initial speed.

By substituting the calculated value of t into the equation, we can find the value of d.

Answer 2

10.9 m

Step-by-step explanation:

We can solve the problem by using the law of conservation of energy.

The initial mechanical energy is just the kinetic energy of the ball:

`E = K_i = (1)/(2)mu^2`

where m is the mass of the ball and u = 16.9 m/s the initial speed.

At a height of h, the total mechanical energy is sum of kinetic energy and gravitational potential energy:

`E=K_f + U_f = (1)/(2)mv^2 + mgh`

where v is the new speed, g is the gravitational acceleration, h is the height of the ball.

Due to the conservation of energy,

`(1)/(2)mu^2 = (1)/(2)mv^2 +mgh\\u^2 = v^2 + 2gh` (1)

Here, at a height of h we want the speed to be 1/2 of the initial speed, so

`v=(1)/(2)u`

So (1) becomes

`u^2 = ((u)/(2))^2+2gh\\(3)/(4)u^2 = 2gh`

So we can find h:

`h=(3u^2)/(8g)=(3(16.9 m/s)^2)/(8(9.8 m/s^2))=10.9 m`