Question

5. When a kid drops a rock off the edge of a cliff, it takes 4.0 s to reach the ground below. When he throws the rock down, it strikes the ground in 3.0 s. What initial speed did he give the rock?

Answer 1

Answer: 11.4m/s

Step-by-step explanation:

kid drops a rock off the edge of a cliff, it takes 4.0 s to reach the ground below

u= 0 , a=9.8 , t= 4 s = ?

s= ut + (1/2) a t^2

s = 0 + (1/2) 9.8 (4)^2

s= 78.4 m

Now for the throw case the initial velocoity is unknow

u = ? , s =78.4 , a=9.8 t=3

s= ut + (1/2) a t^2

(s- (1/2) a t^2 )/ t = u
78.4-(1/2)x9.8x9/3 = u
34.3/3=u
11.4m/s = u

Answer 2

Step-by-step explanation:

assume that the only force acting on the rock is gravity, which gives it a constant acceleration of 9.81 m/s^2.

When the kid drops the rock, it starts with an initial velocity of 0 m/s and falls for a time of 4.0 s before reaching the ground. Using the equation:

d = (1/2)at^2 + vit

where d is the distance fallen, a is the acceleration due to gravity, t is the time, vi is the initial velocity (which is 0), and we solve for a:

a = 2d / t^2

a = 2 * h / t^2 (where h is the height of the cliff)

a = 2 * g * h / t^2 (where g is the acceleration due to gravity)

a = 2 * 9.81 m/s^2 * h / (4.0 s)^2

a = 4.905 h m/s^2

When the kid throws the rock, it starts with an initial velocity vi and falls for a time of 3.0 s before reaching the ground. Using the same equation:

d = (1/2)at^2 + vit