Question

A ball is thrown from ground into air. At a height 9.1m, the velocity is observed to be V=7.6i+6.1j.Then find:
(a)Maximum height,
(b)Horizontal range and
(c)velocity of the ball at the instant of striking the ground.

Answer 1

Given info : A ball is thrown from ground into air. At a height 9.1m, the velocity is observed to be V=7.6 i + 6.1 j

To find : (a) maximum height

(b) horizontal range

(c) velocity of the ball at the instant of striking the ground.

Solution : at maximum height, only horizontal component of velocity appears, vertical component becomes zero.

let ball reaches the maximum height from 9.1 m to x

Then, distance covered from 9.1 to x is s = (x - 9.1)

using formula, v² = u² + 2as

here, u = vertical component of velocity = 6.1 m/s v = 0 , a = - 10 m/s²

So, 0 = (6.1)² + 2(-10)(x - 9.1)

⇒37.21 = 20(x - 9.1)

⇒1.9 + 9.1 = x

⇒x = 11 m

So, maximum height is 11 m

(b) horizontal component of velocity always remains same.

Initial horizontal velocity is 7.6 m/s

now time taken to reach maximum height, t = h/u = 11/7.6 ≈ 1.5 sec

So, time of flight = 2 × t = 3 sec

Now horizontal range = horizontal component of velocity × time of flight

= 7.6 × 3 = 22.8 m

So horizontal range is 22.8 m

(C) at maximum height , velocity of ball, v = (7.6 i + 0 j)

Now, after 1.5 sec ball strikes the ground.

Velocity becomes, v = 7.6 i + (-gt) j

= 7.6 i - 10 × 1.5 j

= 7.6 i - 15 j

Therefore velocity of ball after is (7.6i - 15 j)