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

Byju's Answer

Standard XII

Physics

Periodic Motion

A circketer h...

Question

A circketer hits a ball with a velocity $25$ m/s at $60^{\circ}$ angle to the horizontal. How far above the ground it passes over a fielder $50$ m from the bat (assume the ball is struck very close to the ground)

8.2

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

9.0

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

11.6

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

12.7

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

Open in App

Solution

The correct option is B $8.2$ m
time for ball to cover $50 m$ horizontally
$S = v t$
$50 = 25 c o s 60 t$
$t = 4 s$
height at that time is
$S = u t + 0.5 a t^{2}$
$S = 25 s i n 60 \times 4 - 0.5 \times g \times 16$
$S = 50 \sqrt{3} - 78.4 = 8.2 m$

Suggest Corrections

Similar questions

Q. A cricketer hits a ball with a velocity 25 m/s at

60^{\circ}

above the horizontal. How far above the ground it passes over a fielder 50m from the bat (assume the ball is struck very close to the ground)

A cricketer hits a ball with a velocity 25 m/sat $60^{\circ}$ above the horizontal. How far above the ground it passes over a fielder 50 mfrom the bat (assume the ball is struck very close to the ground)

Q. A cricket ball is hit with velocity

25 m s^{- 1}

60^{o}

above the horizontal. How far above the ground, ball passes over a field

50 m

from the bat(consider the ball is struck very close to the ground)? (Take

\sqrt{3} = 1.7

and

g = 10 m s^{- 1}

)

Q. A cricketer hits a ball from the ground level with a velocity of

(10 \sqrt{2}) m s^{- 1}

at an angle of

45^{\circ}

with the horizontal. What should be the minimum speed of the fielder standing [Along the direction in which ball is hit] at a distance of

5 m

from the batsman so as to catch the ball at ground level. Assume the direction of motion of the ball and the fielder are same.

Q. A ball is thrown from a point

1 m

above the ground. The initial velocity is

20 m / s

at an angle of

40

degrees above the horizontal. Find the maximum height of the ball above the ground. (in

m

)

Join BYJU'S Learning Program

A circketer hits a ball with a velocity 25 m/s at 60∘ angle to the horizontal. How far above the ground it passes over a fielder 50 m from the bat (assume the ball is struck very close to the ground)

The correct option is B 8.2 mtime for ball to cover 50m horizontallyS=vt50=25cos60tt=4sheight at that time is S=ut+0.5at2S=25sin60×4−0.5×g×16S=50√3−78.4=8.2m

A circketer hits a ball with a velocity $25$ m/s at $60^{\circ}$ angle to the horizontal. How far above the ground it passes over a fielder $50$ m from the bat (assume the ball is struck very close to the ground)

The correct option is B $8.2$ m
time for ball to cover $50 m$ horizontally
$S = v t$
$50 = 25 c o s 60 t$
$t = 4 s$
height at that time is
$S = u t + 0.5 a t^{2}$
$S = 25 s i n 60 \times 4 - 0.5 \times g \times 16$
$S = 50 \sqrt{3} - 78.4 = 8.2 m$