What is the equation for: A ball is released from the top of a cliff; it hits th
ID: 1480123 • Letter: W
Question
What is the equation for:A ball is released from the top of a cliff; it hits the ground 4.2 seconds later. Assuming the ball is a free falling object, how many meters is the cliff above the ground below? What is the equation for:
A ball is released from the top of a cliff; it hits the ground 4.2 seconds later. Assuming the ball is a free falling object, how many meters is the cliff above the ground below?
A ball is released from the top of a cliff; it hits the ground 4.2 seconds later. Assuming the ball is a free falling object, how many meters is the cliff above the ground below?
Explanation / Answer
H = Ut + 1/2at^2
=> H = 0 + 1/2gt^2 ( U=0)
=> H = 1/2*9.8*(4.2)^2 = 86.436 m