Question

A ball is gently dropped from the height of 20cm. If its velocity increases

uniformly at the rate of 10m/s2

, with what velocity will it strike the ground?

After what time will it strike the ground?​

Answer 1

Answer:

The ans will be 12 velocity

Answer 2

GIVEN:

• Height from which the ball is dropped, h = 20 cm
• Acceleration, a = 10 m/s²
• Since the ball is dropped gently, its initial velocity is zero. u = 0

TO FIND:

• Velocity of the ball when it strikes the ground, v

• Time at which the ball strikes the ground , t

FORMULAE:

• According to third equation of motion, v² - u² = 2as

• According to the first equation of motion, v = u + at

SOLUTION:

STEP 3: TO CONVERT HEIGHT IN STANDARD UNITS

• SI unit of length is metre, m

• 1 cm = 10^-2 m

• 20 cm = 2 × 10^-2 m

• Height, h = 2 × 10^-2 m

STEP 2: TO FIND VELOCITY

${v}^{2} - {u}^{2} = 2as \\ \\ {v}^{2} - {0}^{2} = 2 \times 10 \times \times 20 \times {10}^{ - 2} \\ \\ {v}^{2} = 400 \times {10}^{ - 2} \\ \\ {v}^{2} = 4 \times 100 \times {10}^{ - 2} \\ \\ {v}^{2} = 4 \times {10}^{2} \times {10}^{ - 2} \\ \\ {v}^{2} = 4 \\ \\ v = \sqrt{4} \\ \\ v = 2 \: \frac{m}{s}$

Velocity of the ball when it strikes the ground is 2 m/s.

STEP 3: TO FIND TIME

$v = u + at \\ \\ 2 = 0 + (10 \times t) \\ \\ 2 = 10t \\ \\ t = \frac{2}{10} \\ \\ t = 0.2 \: seconds$

Time at which the ball strikes the ground is 0.2 seconds.

ANSWER:

• Velocity of the ball when it strikes the ground is 2 m/s.

• Time at which the ball strikes the ground is 0.2 seconds.