You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Brewster's Angle

What is the d...

Question

What is the direction of the ball's velocity just before it hits the ground?

{tan}^{- 1} (3.54)

, below horizontal

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

{tan}^{- 1} (3.54)

, above horizontal

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

{tan}^{- 1} (2.54)

, above horizontal

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

{tan}^{- 1} (2.54)

, below horizontal

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

Open in App

Solution

The correct option is D ${tan}^{- 1} (2.54)$ , above horizontal
Using third kinematics equation:

$v^{2} - u^{2} = 2 a S$

${7.6}^{2} - u_{y}^{2} = - 2 g \times 9.1$

$u_{y}^{2} = 239.76 \Rightarrow u_{y} = 15.5 m / s$

$tan α = \frac{u_{y}}{u_{x}} = \frac{15.5}{6.1} = 2.54$

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

g = 10 m / s^{2}

)

Q. End

A

of a rod

A B

is being pulled on the floor with a constant velocity

v_{0}

as shown. Taking the length of the rod as

l

, at an instant when the rod makes an angle

37^{\circ}

with the horizontal, calculate

The velocity of the CM of the rod

A boy throws a ball with a velocity u at an angle θ with the horizontal. At the same instant he starts running with uniform velocity to catch the ball before it hits the ground. To achieve this, he should run with a velocity of :

Q. If the time of flight of a bullet over a horizontal range

R

T

, then inclination of the direction of projection with the horizontal is

Q. A golfer standing on level ground hits a ball with a velocity of

u = 52 m / s

at an angle

α

above the horizontal. If

tan α = 5 / 12

, then the time for which the ball is at least 15 m above the ground (i.e. between A and B) will be (take

g = 10 m / s^{2}

)

Join BYJU'S Learning Program

What is the direction of the ball's velocity just before it hits the ground?

The correct option is D tan−1(2.54), above horizontalUsing third kinematics equation:v2−u2=2aS7.62−u2y=−2g×9.1u2y=239.76⇒uy=15.5m/stanα=uyux=15.56.1=2.54

The correct option is D ${tan}^{- 1} (2.54)$ , above horizontal
Using third kinematics equation:

$v^{2} - u^{2} = 2 a S$

${7.6}^{2} - u_{y}^{2} = - 2 g \times 9.1$

$u_{y}^{2} = 239.76 \Rightarrow u_{y} = 15.5 m / s$

$tan α = \frac{u_{y}}{u_{x}} = \frac{15.5}{6.1} = 2.54$