Question

A stone is thrown up to a height of 50m with a velocity of 20m/s. After reaching this height, it comes back to the earth. Calculate i) Velocity with which it touches the ground, ii) Time taken to reach this height, iii) Total distance covered, iv) Total time taken. Take g= 10m/s^2.​

Answer 1

EXPLANATION.

A stone is thrown up to a height of 50 m

with a velocity of 20 m/s.

1) = velocity with which it touches

the ground.

$\rm \to \: from \: \: newton \: \: third \: \: equation \: \: of \: \: motion$

$\rm \to \: v {}^{2} = {u}^{2} + 2as$

$\rm \to \: a \: = \pm \: g$

$\rm \to \: a \: = g \: \:(object \: \: move \: \: downward \: )$

$\rm \to \: u \: = 0$

$\rm \to \: v {}^{2} = 0 \: + \: 2 \: \times (g) \times 50$

$\rm \to \: {v}^{2} = 1000$

$\rm \to \: v \: = \sqrt{1000}$

$\rm \to \: v \: = 31.6 \: ms {}^{ - 1}$

2) = Time taken to reach this height.

$\rm \to \: from \: \: newton \: \: first \: \: equation \: \: of \: \: motion$

$\rm \to \: v \: = u \: + \: at$

$\rm \to \: 0 = 20 \: + \: ( - 10)t$

$\rm \to \: t \: = 2 \: \: seconds$

3) = Total distance covered.

$\rm \to \: \: from \: \: newton \: \: second \: \: equation \: \: of \: \: motion$

$\rm \to \: s \: = ut \: + \: \frac{1}{2} at {}^{2}$

$\rm \to \: s \: = \: 20 \times 2 \: + \frac{1}{2} \times ( - 10) \times 2 \times 2$

$\rm \to \: s \: = 40 - 20$

$\rm \to \: s \: = \: 20 \: m$

Total distance covered = 20 + 50 = 70 m

4) = Total time taken.

$\rm \to \: from \: \: newton \: \: second \: \: equation \: \: of \: \: motion$

$\rm \to \: s \: = ut \: + \: \frac{1}{2} at {}^{2}$

$\rm \to \: 70 = 0 + \frac{1}{2} \times 10 \times {t}^{2}$

$\rm \to \: 70 = 5 {t}^{2}$

$\rm \to \: {t}^{2} = 14$

$\rm \to \: t \: = 3.74 \: \: seconds$

Total time = 2 + 3.74 = 5.74 seconds.