if a ball is thrown horizontally from the top of a cliff and takes 3 seconds to reach the beach below. calculate the height of the cliff by considering the vertical motion only.
answer : s=ut+half at(square)
s=0*3+half*10*3
s=3+45
s=48 m/s(square)
a = - 9.8 m/s^2
v = Vi + a t = 0 - 9.8 t
h = Hi + Vi t -(9.8/2) t^2
0 = Hi + 0 (3) - (9.8/2)(3)^2
4.9 (9) = Hi
Hi = 44.1 meters high
To calculate the height of the cliff, we can use the equation for vertical motion:
s = ut + (1/2)at^2
Given:
u = initial vertical velocity (which is 0 since the ball is thrown horizontally)
t = time taken (given as 3 seconds)
a = acceleration due to gravity (which is approximately 10 m/s^2)
Substituting the values into the equation, we have:
s = 0 * 3 + (1/2) * 10 * (3)^2
s = 0 + (1/2) * 10 * 9
s = 0 + 45
s = 45 meters
Therefore, the height of the cliff is 45 meters.
To calculate the height of the cliff using the vertical motion only, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance (height of the cliff)
u = initial velocity (in the vertical direction)
t = time taken to reach the bottom
a = acceleration (due to gravity)
In this scenario, the ball is thrown horizontally, so we can assume that the initial velocity in the vertical direction is 0 m/s. The acceleration due to gravity is approximately 10 m/s^2.
Substituting the given values into the equation, we have:
s = 0 * 3 + (1/2) * 10 * 3^2
s = 0 + (1/2) * 10 * 9
s = 0 + 45
s = 45 meters
Therefore, the height of the cliff is 45 meters.