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

Step-by-step explanation:

t=Time airborne=2.54 s

R= horizontal range= 25.5 m

From the projectile motion equations we know that:

`H=(g*t^(2))/(8)`

`H=(9.8 (m)/(s^(2)) *(2.54s)^(2))/(8)`

H=7.9 m