Question

A projectile is thrown making an angle of 60° with the horizontal with velocity
of 10 m/s. The maximum height attained by the projectile is [Take g = 10
m/s2​

Answer 1

Explanation:

velocity of projectile=u=147 ms

−1

angle of projection α=60

∘

Let the time taken by projectile from O to A, be t and its velocity is v at point A where direction is β=45

∘

. As horizontal component of velocity remains constant during the projectile projectile motion.

⇒vcos45

∘

=ucos60

∘

⇒v×

2

1

=147×

2

1

⇒v=

2

147

ms

−1

For vertical motion,

v

y

=u

y

−gt

⇒vsin45

∘

=45sin60

∘

−9.8t

⇒

2

147

×

2

1

=147×

2

3

−9.8t

9.8t=

2

147

(

3

−1)

t=5.49 s

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