Question

A stone is dropped from a height of 19.6m. Calculate
i) Time taken to fall
ii) How fast does it move at the end of this fall ?
iii) Acceleration after 1 second

Answer 1

Hope it's helpful to u.

Answer 2

Given :

Height (H) = 19.6 m

To Find :

i) Time taken to fall

ii) How fast does it move at the end of this fall ?

iii) Acceleration after 1 second

Solution :

(i) Time taken to fall = $\sqrt{\frac{2H}{g} }$

                                  = $\sqrt{\frac{2 * 19.6}{9.8} }$

                                  = $\sqrt{4}$

                                  = 2 sec

Therefore, the time taken by the stop to fall is 2 sec.

(ii) Using Kinematic equation for uniformly accelerated motion, we have

     v² = u² + 2gH                    (v = Final velocity; u = initial velocity)

⇒    v² = (0)² + 2×9.8×19.6

⇒    v² = 0 + 384.16

⇒    v  = $\sqrt{384.16}$

∴     v  = 19.6 m/sec

Hence, The velocity at the end of the fall is 19.6 m/sec.

(iii) The acceleration will be constant and will be equal to acceleration due to gravity (g). Thus, the acceleration after 1 sec will also be g i.e, 9.8 m/sec².