A stone is projected from the ground with a velocity of $25 m / s$ . $2$ seconds later, it just clears wall $5 m$ high. Then angle of projection of the stone and the greatest height reached are- $[$ Neglect air resistance, Assume $g = 10 m / s e c^{2}$ $]$

30

7.8 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

60

8.7 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

30

8.7 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

60

8.7 c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $30$ , $7.8 m$
We know that $s = u t + 0.5 a t^{2}$
$⟹ 5 = 25 s i n θ \times 2 - 0.5 \times g \times 4$

$∴ s i n θ = \frac{1}{2} ⟹ θ = 30^{o}$

and $H = \frac{25^{2} s i n^{2} 30}{20} = 7.8 m$

Suggest Corrections

Similar questions

A stone is projected from the ground with velocity 25 m/s. Two seconds later, it just clears a wall 5 m high. The angle of projection of the stone is $(g = 10 m / s e c^{2})$

A stone is projected from the ground with a velocity of 25 m/s. Two seconds later, it just clears a pole of 5m height. The angle of projection is

Q. A stone is projected the ground with velocity 25 m/s. Two seconds later, it just clears a wall 5 m high. The angle of projection of the stone is

(g = 10 m / {s e c}^{2})

Q. A stone is projected from the ground with a velocity of

14 m s^{- 1}

one second later it clean a wall

2 m

high. The angle of projection is

(g = 10 m s^{- 2})

Join BYJU'S Learning Program

A stone is projected from the ground with a velocity of 25m/s. 2 seconds later, it just clears wall 5m high. Then angle of projection of the stone and the greatest height reached are- [Neglect air resistance, Assume g=10m/sec2]

The correct option is A 30, 7.8mWe know that s=ut+0.5at2⟹5=25sinθ×2−0.5×g×4∴sinθ=12⟹θ=30oand H=252sin23020=7.8m

A stone is projected from the ground with a velocity of $25 m / s$ . $2$ seconds later, it just clears wall $5 m$ high. Then angle of projection of the stone and the greatest height reached are- $[$ Neglect air resistance, Assume $g = 10 m / s e c^{2}$ $]$

The correct option is A $30$ , $7.8 m$
We know that $s = u t + 0.5 a t^{2}$
$⟹ 5 = 25 s i n θ \times 2 - 0.5 \times g \times 4$

$∴ s i n θ = \frac{1}{2} ⟹ θ = 30^{o}$

and $H = \frac{25^{2} s i n^{2} 30}{20} = 7.8 m$