Question

helppppppp pleaseeeeeeeee

Answer 1

8.9 m/s

Explanation:

Note that the jumper launched from and then landed also on the ground. This means that the jumper's vertical displacement y is zero. From this information, we can determine the amount of time it took for him to make the jump using the equation

$y = v_{0y}t - \frac{1}{2}gt^2 = (v_0\sin{45°})t - \frac{1}{2}gt^2$

Since y = 0, then the equation above becomes

$0 = (v_0\sin{45°})t - \frac{1}{2}gt^2$

Solving for t, we get

$t = \dfrac{2v_0\sin{45°}}{g}$

We also know that

$x = v_{0x}t = (v_0\cos{45°})t$

Using the expression for t we got earlier into the equation above, we get

$x = (v_0\cos{45°})\left(\dfrac{2v_0\sin{45°}}{g}\right)$

$\;\;\;\;=\dfrac{v_0^2}{g}(2\sin{45°}\cos{45°})$

$\;\;\;\;= \dfrac{v_0^2}{g}\sin{2(45°)°}$

$\Rightarrow x = \dfrac{v_0^2}{g}$

Solving for $v_0,$ we get

$v_0^2 = gx \Rightarrow v_0 = \sqrt{gx}$

Putting in the given values, we find that the final velocity is

$v_0 = \sqrt{(9.8\:\text{m/s}^2)(8.0\:\text{m})} = 8.9\:\text{m/s}$

Note: You can also solve this problem by using the range equation:

$R = \dfrac{v_0^2}{g}\sin{2\theta}$