We can use the equation of motion to solve this problem. The equation is:
y = y0 + v0yt + 0.5at^2
Where:
y = final vertical displacement = 0 (since the rock reaches the ground)
y0 = initial vertical displacement = 25 m
v0y = initial vertical velocity = 0 (since the rock is thrown horizontally)
a = acceleration due to gravity = -9.8 m/s^2 (taking downward direction as negative)
t = time taken to reach the ground
Substituting the values into the equation, we get:
0 = 25 + 0 + 0.5*(-9.8)*t^2
0 = 25 - 4.9t^2
4.9t^2 = 25
t^2 = 25/4.9
t^2 = 5.102
t = sqrt(5.102)
t ≈ 2.26 seconds
Therefore, it takes approximately 2.26 seconds for the rock to reach the horizontal ground below.
A rock is thrown horizontally with a speed of 20 m/s from a vertical cliff of height 25 m.
How long does it take the rock to reach the horizontal ground below?
1 answer