Question

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 28.0 ∘ above the horizontal. You can ignore air resistance.

Answer 1

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,

$v^2-u^2=2ay$

At maximum height, v = 0

Here, a = -g

$-(u\ sin\theta)^2=-2gy$

$(u\ sin\theta)^2=2gy$

$y=\dfrac{(u\ sin\theta)^2}{2g}$

$y=\dfrac{(26\times \ sin(28))^2}{2\times 9.8}$

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.