Question

6.A rock is dropped from the top of a 35 m high building.
a. How long does it take to reach the ground?
b. what is the speed of the rock as it strikes the ground? (g=9.8m/s2)​

Answer 1

Explanation:

Using equations of motion: S = ut + ½ at²

Here, a = g = 9.8 ; S = 25, since it is dropped(not thrown with force) u = 0.

=> 35 = (0)t + ½ (9.8)t²

=> 35 = (4.9)t²

=> 35/4.9 = t²

=> 2.67 ≈ t, it takes 2.67 sec to reach the ground.

Using v = u + at,

=> v = 0 + (9.8)(2.67)

=> v = 26.19 m/s

speed of the rock as it strikes the ground is 26.19 m/s

Answer 2

Height of the building, h = 35 m

Initial speed with which it is dropped, u = 0 m/s

Acceleration due to gravity, g = 9.8 m/s²

a. Time taken by the rock to reach the ground = t

Now, h = ut + $\dfrac{1}{2}$at²

• 35 = $\dfrac{1}{2}$ × 9.8 × t²

• t² = $\dfrac{35 × 2}{9.8}$

• t² = 7.14285

• t = $\sqrt{7.14285}$

• t = 2.67 s

b. Final speed of the rock = v

• v = u + at

• v = 0 + 9.8 × 2.67

• v = 26.19 m/s

$\bold{Hope\;it \; helps\;!}$