A ball thrown from the edge of a 35m high cliff takes 3.5s to reach the ground below. What was the ball's initial velocity?
Try X = Vo + (1/2)at^2.
Solve for Vo.
To find the ball's initial velocity, we can use the following equation of motion:
h = (1/2) * g * t^2
where:
h is the height of the cliff (35m),
g is the acceleration due to gravity (9.8 m/s^2), and
t is the time taken for the ball to reach the ground (3.5s).
First, we can rearrange the equation to solve for the initial velocity (u):
u = sqrt(2gh)
Let's plug in the values and calculate the initial velocity of the ball:
u = sqrt(2 * 9.8 * 35)
u = sqrt(686)
u ≈ 26.2
Therefore, the ball's initial velocity was approximately 26.2 m/s.
To find the ball's initial velocity, we can use the equations of motion and the principle of conservation of energy.
First, let's assume that the positive direction is upward. The acceleration due to gravity is -9.8 m/s², which means it is in the opposite direction to the positive direction we have assumed. We will also assume that there is no air resistance.
The equation we'll use to solve for the initial velocity is:
v = u + at
Where:
v = final velocity (which is 0 m/s because the ball hits the ground and stops)
u = initial velocity (what we are trying to find)
a = acceleration (-9.8 m/s²)
t = time taken (3.5 s)
Rearranging the equation, we have:
u = (v - at)
Substituting the values we know:
u = (0 - (-9.8 * 3.5))
u = 0 + 34.3
u = 34.3 m/s
So, the ball's initial velocity was 34.3 m/s.