a rock is thrown 15 m/s . find the height after 2 s .
Answers
Explanation:
Because the rock is experiencing a constant acceleration downwards at 9.81 m/s^2, the constant acceleration formulas come into play.
The initial velocity has been given, 15 m/s.
The initial angle? 90 degrees. It is being thrown upwards, after all.
So, how to find the height? Very simple.
s=ut+ 1/2at^2
s=(15)(2)+1/2(-9.81)(2)^2
s=10.38
The rock is 10.38m in the air.
Minor edit: The initial height is pretty obvious. If it was thrown by a person, it was obviously in the person’s hand, and the height of the person’s hand abovd the ground is rather negligible and can be ignored. In the case that you really want to include the height of the person’s hand above the ground,
Assuming the height the person’s hand above the ground at the time the rock left the person’s hand was 1.5m,
10.38m+1.5m= 11.88m
It really doesn’t affect the final result by that much.
Answer:
h=10 m
Explanation:
u=15 m/s
a=-g=-10 m/s²
t=2 s
h=?
S=ut-1/2gt²
=15×2-1/2×10×2²
=30-20=10 m
=>h=10 m