Question

Work out? If you throw a maximum range of 45m, what maximum height you might jump assuming the same initial speed in both cases? Thanks

Answer 1

22.5 m

Explanation:

In order to throw at a maximum range, we throw at a `45^circ` angle. We can use projectile motion equations to figure out the velocity from the range and angle:

`y(t) = v_y t -1/2 g t^2`
`x(t) = v_x t`

We also know that `v_x = v_y` because it's at a `45^circ` angle. Therefore, when the object hits the ground,
`y(t) = 0` if `t = 0` (when we started) or when `t = (2v_y)/g`.

Plugging this into the equation for `x(t)`,
`x = v_x (2v_y)/g = (2v_x^2)/g = R`

where `R` is our range.

`v_x = sqrt((gR)/2) approx sqrt(1/2 * (10ms^-2) * 45 m) = 15 ms^-1`

Therefore, if we have a y and x velocity of 15, our total speed is `15ms^-1 * sqrt(2) approx 21 ms^-1`.

If you jump at that speed, your max height will be

`h = (v^2)/(2g) = (2 * (gR) / 2)/(2g) = R/2`

So after all that, it's just half of your range, or 22.5 m.