Question

Ann successfully jumped over a 25.5 m-wideriver. Assuming that she started and landed at the same level and was airborne for 2.54 s, what height from her starting point did our extreme river jumper achieve?

Answer 1

Answer:

Explanation:

t=Time airborne=2.54 s

R= horizontal range= 25.5 m

From the projectile motion equations we know that:

$H=\frac{g*t^{2}}{8}$

$H=\frac{9.8 \frac{m}{s^{2}} *(2.54s)^{2}}{8}$

H=7.9 m