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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

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9.0

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11.6

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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
$S = u t + 0.5 a t^{2}$
$S = 25 s i n 60 \times 4 - 0.5 \times g \times 16$
$S = 50 \sqrt{3} - 78.4 = 8.2 m$

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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
$S = u t + 0.5 a t^{2}$
$S = 25 s i n 60 \times 4 - 0.5 \times g \times 16$
$S = 50 \sqrt{3} - 78.4 = 8.2 m$