You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Body in Water

A particle is...

Question

A particle is thrown horizontally with $20 m / s$ from the top of a tower of height $20 m (g = 10 m / s^{2})$ . Angle with horizontal at which it will hit the ground is:-

30^{\circ}

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

60^{\circ}

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

53^{\circ}

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

45^{\circ}

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

Open in App

Solution

The correct option is D $45^{\circ}$
$(D)$
$v_{x} = u_{x} = 20$
using: $v^{2} - u^{2} = 2 a s$ to find $v_{y}$
$v_{y} = \sqrt{2 g h}$
$v_{y} = 20$
$t a n θ = \frac{v_{y}}{v_{x}}$
$t a n θ = 20 / 20 = 1$
$θ = 45^{o}$

Suggest Corrections

Similar questions

Q. A body is projected from the top of a tower of height

32 m

with a velocity of

20 m/s

at an angle of

37^{\circ}

above the horizontal. Find the horizontal distance travelled before it hits the ground.
(Take

g = 10 m / s^{2}

)

Q. A ball is thrown up from the top of a tower with an initial velocity of

10

m/s at an angle of

30^{o}

with the horizontal. It hits the ground at a distance of

17.3

m from the base of tower. Calculate the height of the tower.

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

Q. Two particles

A

and

B

are projected simultaneously from the top of two towers of height

10 m

and

20 m

respectively. Particle

A

is projected upwards at an angle of

45^{\circ}

with horizontal at a speed of

10 \sqrt{2} m/s

and particle

B

is projected horizontally at a speed of

10 m/s

towards

A

. What is the distance between towers if particles collide in the air ?Take

g = 10 {m/s}^{2}

Q. A body is projected from the top of a tower of height

32 m

with a velocity of

20 m/s

at an angle of

37^{\circ}

above the horizontal. Find the angle below the horizontal with which it hits the ground.
(Take

g = 10 m / s^{2}

)

Q. A particle is projected from the ground at an angle of

60^{\circ}

with horizontal with speed

u = 20 m / s

. The radius of curvature of the path of the particle, when its velocity makes an angle of

30^{\circ}

with horizontal is
(Take

g = 10 m / s^{2}

)

Join BYJU'S Learning Program

A particle is thrown horizontally with 20m/s from the top of a tower of height 20m(g=10m/s2). Angle with horizontal at which it will hit the ground is:-

The correct option is D 45∘(D)vx=ux=20using: v2−u2=2as to find vyvy=√2ghvy=20tanθ=vyvxtanθ=20/20=1θ=45o

A particle is thrown horizontally with $20 m / s$ from the top of a tower of height $20 m (g = 10 m / s^{2})$ . Angle with horizontal at which it will hit the ground is:-

The correct option is D $45^{\circ}$
$(D)$
$v_{x} = u_{x} = 20$
using: $v^{2} - u^{2} = 2 a s$ to find $v_{y}$
$v_{y} = \sqrt{2 g h}$
$v_{y} = 20$
$t a n θ = \frac{v_{y}}{v_{x}}$
$t a n θ = 20 / 20 = 1$
$θ = 45^{o}$