Math, asked by eixeralobba, 14 hours ago

A ball is thrown straight up at an initial velocity of 54 feet per second. The height of the ball (t) seconds after it is thrown is given by the formula h(t) = 54t - 12t²

How many seconds after the ball is thrown will it return to the ground?

Answers

Answered by Anonymous
4

thrown straight up at an initial velocity of 54 feet per second. The height of the ball (t) seconds after it is thrown is given by the formula h(t) = 54t - 12t²

Answered by venomgirl8
2

Answer:

A ball is thrown straight up at an initial velocity of 54 ft/sec. How many seconds after the ball is throw will it return to the ground?

A couple of ways to get this. This question must assume that the ball is thrown up from the ground, so that the height it returns to is the same as the starting point.

v = vi + at

Calculate t to maximum height when v = 0, then double the time, since time up equal time down.

Alternatively, you should know that v = -vi, when the object returns to its original throwing height. So this means v = -54 ft/s, and the total time can be calculated directly.

Since the starting point has not been mentioned, simply assume that it start at the ground. Let the upward direction be positive.

Then we have the initial velocity u ft/s = 54 ft/s and

acceleration a ft/(s^2) = - 32 ft/(s^2)

use the equation s = ut + (1/2)a (t^2)

when return to the ground, s ft= 0 ft

substitute all those figures into the equation, we have

0 = 54t - 16 t^2

or 2t(27 - 6t) = 0

we have t = 0 (the initial state) or t = 4.5

It took 4.5 s to return to the ground.

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