#### Question

# Math help? Ball dropped from a height of 70 meters?

David drops a ball from a bridge at an initial height of 70 meters.

a.)What is the height of the ball to the nearest tenth of a meter exactly 2 seconds after he releases the ball?

b.)How many seconds after the ball is released will it hit the ground?

Please and Thank you. I don't understand how to solve it.

#### Answers

The position function for a free-falling object is given by

h(t) = - 4.88t2 + v0t + h0

where h = height in meters, t = time in secs., v0 = initial velocity in m./sec. and h0 = initial height in meters.

v0 = 0

h0 = 70 m

Position Function for this Problem:

h(t) = - 4.88t2 + 70

a.) t = 2 secs.

h(2) = - 4.88(2)2 + 70

h(2) = - 4.88(4) + 70

h(2) = -19.52 + 70

h(2) = 50.5

Height of Ball at 2 secs. = 50.5 meters

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b.) Time to Impact: h(t) = 0:

- 4.88t2 + 70 = 0

- 4.88t2 = - 70

t2 = - 70 / - 4.88

t2 = 14.34

t = √14.24

t = 3.8

Time to Impact = 3.8 secs.

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This is a physics problem. The ball is released with an initial velocity of 0 and accelerates at 9.81 m/s2 until it hits the ground. That should be enough to get you started.