Question

a person throws a ball with 40 under root 2m/sec at angle 45 degree from horizontal. Then ball travels, in projectile path. (Consider g = 10m/s square).

Find out time of flight.

CBSE Physics class 11

Answer 1

Answer:

Explanation:

The total time of flight is

Resultant displacement is zero in Vertical direction.

Therefore, by using equation of motion

 s=ut−21​gt2

gt=2sinθ

t=g2sinθ​

(b). The  horizontal range is

Horizontal range OA = horizontal component of velocity × total flight time

 R=ucosθ×g2usinθ​

R=gu2sin2θ​

(c). The maximum height is

It is the highest point of the trajectory point A. When the ball is at point A, the vertical component of the velocity will be zero.

By using equation of motion

 v2=u2−2as

0=u2sin2θ−2gH

H=2gu2sin2θ​

Answer 2

Here,

u = 40√2 m/s

∅ = 45°

g = 10 m/s²

Then,

$T = \frac{2u \: sin∅}{g} = \frac{2(40 \sqrt{2}) \: sin {45}^{o} }{10} = \frac{80 \sqrt{2} }{10 \sqrt{2} } \\ \\ = 8 \: seconds$

Hope It Helps:)

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