Answer:
The maximum height above the roof that the rock reaches is 7.6 meters.
Explanation:
Given that, a man stands on the roof of a building of height 14.0 m and throws a rock with a velocity of magnitude 26.0 m/s at an angle of 28 degrees above the horizontal.
It is assumed to find the maximum height above the roof that the rock reaches. Let it is given by y. So,
[tex]v^2-u^2=2ay[/tex]
At maximum height, v = 0
Here, a = -g
[tex]-(u\ sin\theta)^2=-2gy[/tex]
[tex](u\ sin\theta)^2=2gy[/tex]
[tex]y=\dfrac{(u\ sin\theta)^2}{2g}[/tex]
[tex]y=\dfrac{(26\times \ sin(28))^2}{2\times 9.8}[/tex]
y = 7.60 meters
So, the maximum height above the roof that the rock reaches is 7.6 meters. Hence, this is the required solution.
