Answer:
A.) 5s
B.) 2.29s
C.) 39.6 m/s
Step-by-step explanation:
Given that a kid throws a rock off the edge of a cliff 60.0 m high. When he drops the rock, it takes 5,0sec for it to reach the ground below.
a.) Determine the time it takes to reach the ground below,
The time is 5s
b.) How long would it take the rock to hit the ground if he threw it down with a speed of 15m/sec?
Using the second equation of motion:
H = Ut + 1/2gt^2
Substitute all the parameters into the formula
60 = 15t + 1/2 × 9.8 × t^2
60 = 15t + 4.9t^2
Multiply all by 10
49t^2 + 150t - 600
t^2 + 3.06t = 12.24
( t + 1.53)^2 = 12.24 + 2.3409
t + 1.53 = sqrt ( 14.5809)
t = 3.82 - 1.53
t = 2.29s
is hit stmight upward and reach a maximum height of 80.0 m.
Using the third equation of motion
V^2 = U^2 - 2gH
At maximum height, V = 0
0 = U^2 - 2 × 9.8 × 80
U^2 = 1568
U = sqrt ( 1568)
U = 39.6 m/s