Question

A particle is thrown with a velocity u m/s at an angle of to the horizon. After what time will the particle
be moving at right angle to its initial direction?
pls solve the question with explanation else don't ​

Answer 1

Answer:

making an angle θ with the vertical. It just crosses the top of two poles each of height h after 1s and 3s respectively. The maximum height of projectile is-

question

December 27, 2019avatar

Lina Srivastav

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ANSWER

times taken to reach P is 1s therefore

v

2

=u

2

+2aS

v

2

=u

2

cos

2

θ−2gh

let the distance from P to reach at the top be x

v

2

=u

2

+2as

0=u

2

cos

2

θ−2gh−2gx

x=

2g

u

2

cos

2

θ

−h

time taken is v=u+at

0=ucosθ−g−gt

t=

g

ucosθ

−1

time taken to reach top from P is 1s

2g=ucosθ

maximum height H=

2g

u

2

cos

2

θ

as the verical component is ucosθ

therefore H=

2g

(2g)

2

=2g