Question

A diver jumps from a height of 10m to a pool. If his mass is 70kg nd the speed when he is 5m above the water surface.(Take g = 10m/s2)

Answer 1

We can solve this problem by using mechanical energy conservation theorem. Which is,

Kinetic energy of the body + Potential energy of the body = constant

$E_{K.E} +E_{P.E}  = Constant$

$(1/2)mV^{2} + mgh = Constant$------- (1)

Where,

m = mass of the body

V = velocity of the body

g = acceleration due to gravity

h = height from the potential zero point

• If a driver is in initial position,

m = 70 kg

V = 0

g = 10 m/s2

h = 10m

Using equation (1),

$(1/2)*70*0 + 70*10*10 = constant\\7000 = constant$---- (2)

• When he is 5m above the water surface,

m = 70 kg

V = U

g = 10 m/s2

h = 5m

Using equation (1),

$(1/2)*70*V^{2} + 70*10*5 = constant\\35V^{2} + 3500 = constant$------- (3)

by equaling equation (2) & (3),

$7000 = 35V^{2} + 3500\\35V^{2} = 3500\\V^{2} = 100\\V = 10 m/s$

Answer : 10 m/s

Answer 2

Initial velocity of the body = 0

Height at the diver jumps = 10 m.

Height at which velocity is to find = 5 m.

Therefore, distance covered  = 10 - 5 = 5 m.

Acceleration (g) = 10 m/s².

Using the formula,

v² - u² = 2aS

∴ v² = 2 × 10 × 5

∴ v² = 100

∴  v = 10 m/s.

Hence, velocity of the diver at the height of 5 m is 10 m/s.

Hope it helps.