Question

A ball is thrown upward at a speed v0 at an angle of 58.0˚ above the horizontal. It reaches a maximum height of 8.0 m. How high would this ball go if it were thrown straight upward at speed v0?

Answer 1

Answer:

11.245 m

Explanation:

The vertical component of the initial velocity v0 is

$v_v = v_0sin58^0 = 0.848v_0$

This makes the ball reach a maximum height of 8m. If we apply the conservation law of mechanical energy, its kinetic energy is converted to potential energy when it travels to the maximum height

$E_p = E_k$

$mgh = mv_v^2/2$

where m is the mass and h = 8 m is the maximum vertical distance traveled, g = 9.81m/s2 is the gravitational acceleration

we can divide both sides by m

$gh = v_v^2/2$

$(0.848v_0)^2 = 2gh = 2*9.81*8 = 156.96$

$0.848v_0 = \sqrt{156.96} = 12.53$

$v_0 = 12.53 / 0.848 = 14.77 m/s$

So if the ball is directed fully upward at v0 speed then we can apply the same equation to find the new H

$E_p = E_k$

$mgH = mv_0^2/2$

$H = \frac{v_0^2}{2g} = \frac{14.77^2}{2*9.81} = \frac{218.3}{19.62} = 11.245m$