Question

A ball rolls off a 1.00 m high table and lands 1.60 m along the floor. Find (a) its initial speed and (b) the time of flight.

Answer 1

Solution) Given height , H = 1.00 m

Range , X = 1.60 m

(a) initial speed , V = ?

We have

H = (Vy)t + (1/2)(g)(t^2)

Vy = 0 ( initially ball is moving in horizontal direction so vertical velocity Vy = 0 )

H = (1/2)(g)(t^2)

2H = g(t^2)

t^2 = (2H)/(g)

t = ((2H)/(g))^(1/2)

t = ((2×1)/(9.8))^(1/2)

t = ((2)/(9.8))^(1/2)

t = 0.45 s

Along horizontal direction , X = Vt

V = (X)/(t)

V = (1.60)/(0.45)

V = 3.55 m/s

(b) time of flight , t = ?

Time of flight is time it remains in air .

And is same as solved in part(a)

t = ((2H)/(g))^(1/2)

t = ((2×1)/(9.8))^(1/2)

t = 0.45 s