Answer:
s = 2.79 m
Explanation:
Given that,
A ball have a speed of 7.4 m/s just before it hits the ground.
Initial velocity of the ball was 0 (at rest)
We need to find the height from where the ball was dropped. It means we need to find the distance covered by it. Let it is h.
Using the third equation of motion to find it as follows :
[tex]v^2-u^2=2as\\\\s=\dfrac{v^2}{2g}\\\\s=\dfrac{(7.4)^2}{2\times 9.8}\\\\s=2.79\ m[/tex]
So, the ball is dropped from a height of 2.79 m.
The height from which you would have to drop a ball so that it would have a speed of 7.4 m/s just before it hits the ground is 2.79m
According to the third equation of motion;
[tex]v^2=u^2+2as[/tex]
where;
v is the final velocity = 7.4m/s
Initial velocity = 0m/s
s is the distance
a = g= 9.8m/s²
Substitute the values into the formula to have:
[tex]7.4^2=0^2+2(9.8)s\\54.76 + 19.6s\\s=\frac{54.76}{19.6}\\ s=2.79m[/tex]
Hence the height from which you would have to drop a ball so that it would have a speed of 7.4 m/s just before it hits the ground is 2.79m
Learn more on equation of motion here: https://brainly.com/question/25951773
