Question

A body is projected horizontally from the top of a building of height'h', velocity of projection i u Find: (1 the time it will take to reach the ground. (2) velocity with which the body reaches the ground.

Answer 1

Horizontal projectile motion.

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Answer 2

Answer:

A body reaches the ground with the velocity of $\sqrt{2gh}$ at time $\sqrt{\dfrac{2h}{g}}$.

Explanation:

Given that,

Height = h

Projection velocity =u

(I). We need to calculate the time it will take to reach the ground

Using second equation of motion

$s_{y}=u_{y}t+\dfrac{1}{2}gt^2$

Where, s = height

u = velocity

g = acceleration due to gravity

t = time

Put the value in the equation

$h=0+\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2h}{g}}$

(II). We need to calculate the velocity with which the body reaches the ground

Using first equation of motion

$v_{y} = u_{y}+g_{y}t$

$v_{y}=0+g\times\sqrt{\dfrac{2h}{g}}$

$v_{y}=\sqrt{2gh}$

Hence, A body reaches the ground with the velocity of $\sqrt{2gh}$ at time $\sqrt{\dfrac{2h}{g}}$.