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?

50.4 m   is the height of the ball to the nearest tenth of a meter exactly 2 seconds after he releases the ball

3.78 seconds after the ball is released will  hit the ground

Step-by-step explanation:

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

To Find:  (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?

Solution : Let y(t) denote the number of meters above the ground the ball is after t seconds,

Using equation of motion :

y(t) = 70 - ½gt²  

Using g = 9.8 m/sec²,

we have  y(t) = 70 - 4.9t²   ----a

Thus (a)  y(2) = 70 - 4.9 * 2² = 70 - 19.6 = 50.4 m  

(b)  how many seconds after the ball is released will it hit the ground

When ball hits the ground y will be 0

Putting value in equation a  

0 = 70 - 4.9t²  

Thus t ≈ 3.78 sec

Hence 50.4 m   is the height of the ball to the nearest tenth of a meter exactly 2 seconds after he releases the ball

3.78 seconds after the ball is released will  hit the ground

70

step-by-step explanation:

2*5=10

4*15=60

60+10=

70

step-by-step explanation:

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