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

Byju's Answer

Standard XII

Physics

Summary for Time, Height and Range

A boy playing...

Question

A boy playing on the roof of a $10$ building throws a ball with a speed of $10$ at an angle of $30$ with the

7.56m

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

8.66m

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

9.86m

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

5.67m

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

Open in App

Solution

The correct option is A 8.66m
$R = \frac{u^{2} s i n 2 θ}{g}$

$R = \frac{10^{2} \times s i n (2 \times 30)}{10}$

$R = \frac{10 \sqrt{3}}{2} = 5 \sqrt{3} = 8.66 m$

Suggest Corrections

Similar questions

Q. A boy playing on the roof of a

10

m high building throws a ball with a speed of

10

m s^{- 1}

at an angle of

30^{o}

with the horizontal. How far from the throwing point with the ball be at the height of

10

m from the ground?

(g = 10 m s^{- 2}, sin 30^{o} = 1 / 2, cos 30^{o} =

Q. A boy playing on the roof of a

10 m

high building throws a ball with a speed of

10 m / s

at an angle of

30^{\circ}

with the horizontal. The time taken by the ball to cross the point which is at the height of

10 m

from the ground is

Q. A boy playing on the roof of a

10 m

high building throws a ball with a speed of

10 m / s

at an angle of

30^{o}

with horizontal. How far from the throwing point will the ball be at the height of

10 m

from the ground ?

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

A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at an angle of $30^{\circ}$ with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground $(g = 10 m / s^{2}, s i n 30^{\circ} = \frac{1}{2}, c o s 30^{\circ} = \frac{\sqrt{3}}{2})$

Join BYJU'S Learning Program

A boy playing on the roof of a 10 building throws a ball with a speed of 10 at an angle of 30 with the

The correct option is A 8.66mR=u2sin2θgR=102×sin(2×30)10R=10√32=5√3=8.66m

A boy playing on the roof of a $10$ building throws a ball with a speed of $10$ at an angle of $30$ with the

The correct option is A 8.66m
$R = \frac{u^{2} s i n 2 θ}{g}$

$R = \frac{10^{2} \times s i n (2 \times 30)}{10}$

$R = \frac{10 \sqrt{3}}{2} = 5 \sqrt{3} = 8.66 m$