Answer:
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
[tex]\frac{70}{4.9}= t^{2}[/tex]
[tex]14.2857= t^{2}[/tex]
[tex]\sqrt{14.2857} =t[/tex]
[tex]3.779=t[/tex]
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
