Question

A body is thrown vertically upward from the surface of earth and reaches a maximum height of 500m . The magnitude of initial velocity of the body is[Take g=10m/s^2

Answer 1

u =0m/s

v=???

h=500m.

g=10m/s^2

Using 3 equation of motion.

v^2 = u^2 - 2gh

v^2 = 0^2 - 2×10×500

V^2 = 20×500.

V^2 = 10000.

v = 100m/s.

Answer 2

Answer:

The magnitude of the initial velocity of the body would be equal to 100ms⁻¹.

Explanation:

Consider that $u$ is the initial velocity of the body,

When the body is thrown upward in vertical direction, at highest point its velocity will be final velocity, v=0.

The maximum height body reached, $h=500m$

The body falls back under gravity ,$a=g=-10ms^{-2}$

where -ve sign because body falls against the gravity.

From 3rd equation of motion:

$v^{2}-u^{2}=2aS$                                                     ............(1)

The equation (1) becomes,

$v^{2}-u^{2}=2gh$                                                      ..............(2)

Put the value of v, g and h  in eq.(2),

$(0)^{2}-u^{2}=2(-10)(500)\\ u^{2}=10000\\u=\sqrt{10000} \\u=100ms^{-1}$

Hence the body thrown upward vertically, it has initial velocity will be 100ms⁻¹.