Question

1.A stone dropped from a window reaches the ground in 0.5 seconds (given g = 10m/s^2).
(i)Calculate the speed just before it hits the ground.
(ii)What is the average speed at t= 0.5 s?
(iii)Calculate the height of window from the ground

Answer 1

We have,

Time to reach ground = ½ s

Value of ‘g’ = 10 m/s²

(i) We know,

$\boxed{\tt{\star \ v = u + gt} }$

⇒ v = gt = (10 m/s²)(½ s) = 5 m/s (1).

(ii) We know,

$\boxed{\tt{\star \ v_{av} = \frac{1}{2}(v + u)}}$

⇒ v(av) = (5 m/s + 0 m/s)/(2) = 5/2 m/s

⇒ v(av) = 2.5 m/s (2).

(iii) We know,

$\tt{\boxed{\tt{ \star \ ∆h = ut + \frac{1}{2} gt^2}} }$

⇒ ∆h = ½ gt²

⇒ ∆h = ½(10 m/s²)(0.5 s)²

⇒ ∆h = (5 m/s²)(0.25 s²)

⇒ ∆h = 1.25 m (3).