A projectile is fired horizontally with a velocity of 48 m/s from the hill 294 m high. Find the time taken to reach the ground. A. 7.75 s
B. 10.0 s
C. 60.5 s
D. 15.2 s

Using only vertical components to substitute into the kinematic equation d=Vi(t)+1/2(a)(t)^2, we have:

d=294m
Vi= 0m/s
Vf=?
a=10m/s^ (use 9.81 in substitution for regents physics)
t=?

First let’s solve for Vf using Vf^2=Vi^2+2(a)(d)

Vf^2= (0m/s) + 2(10m/s^2)(294m)
[RAD]Vf^2= [RAD]5880m/s
Vf=76.7m/s

Now we can solve for time using d=Vi(t)+1/2(a)(t)^2

294m=(0m/s)+1/2(10m/s^2)(t)^2
(294m)/5=((5m/s)(t)^2)/5
[RAD]t^2= [RAD]58.8sec
t=7.66 sec

*when using 9.81m/s^2 as (a) value, the answer is 7.75 sec

This is the answer: body 1 + engine 2