Question

A ball is projected from ground at angle 8 with the horizontal. After t = 1 s, it is moving at 45° with the horizontal and after t = 2 second, it is moving horizontally. What is speed of projection of ball? [g = 10 m s -²]

1) 10√2 m/s
2) 10√3 m/s
3) 20 m/s
4) 10√5 m/s​

Answer 1

Correct Question :

A ball is projected from ground at angle 0 with the horizontal. After t = 1 s, it is moving at 45° with the horizontal and after t = 2 second, it is moving horizontally. What is speed of projection of ball? [g = 10 m s -²]

Answer :

Let the tangent make an angle α with the horizontal.

For this motion under gravity,

$\bf\tan\alpha=\dfrac{u\ \sin\theta-gt}{u\ \cos\theta}$

where,

• u = speed

Case 1 :

When,

• α = 45°
• t = 1 s

$\implies\sf\tan45\degree=\dfrac{u\ \sin\theta-gt}{u\ \cos\theta}$

$\\ \implies\sf1=\dfrac{u\ \sin\theta-gt}{u\ \cos\theta}$

$\\ \implies\sf u\ \cos\theta=u\ \sin\theta-gt\ \ \bf------- (i)$

Case 2 :

When the ball is moving horizontally,

• α = 0°
• t = 2 s

$\implies\sf\tan0\degree=\dfrac{u\ \sin\theta-gt}{u\ \cos\theta}$

$\\ \implies\sf0=\dfrac{u\ \sin\theta-gt}{u\ \cos\theta}$

$\\ \implies\sf u\ \sin\theta-gt=0\ \ \bf------- (ii)$

From (ii),

$\sf u\ \cos\theta=2g-g=g$

$\\ \sf u \cos\theta=g\ \ \bf------- (iii)$

Squaring and adding eq. (ii) and (iii),

$\\ \sf u^2(\sin^2\theta+\cos^2\theta=5g^2$

$\\ \sf u=\sqrt{5g^2}$

$\\ \sf u=g\sqrt{5}$

• g = 10 m/s²

$\\ \sf u=10\sqrt{5}$

$\\ {\boxed{\bf u=10\sqrt{5}\ m/s.}}$

The speed of ball is 10√5 m/s.

The correct option is (4). ✔