Question

A particle is projected with a velocity of 10 m/sec at an angle of

elevation 60°. Determine : 8

i) Equation of its path

ii) Horizontal range

iii) Time of flight​

Answer 1

Answer:

particle is projected with velocity 10m/s at an angle of 60° from ground. Then the vertical component of velocity vector when instantaneous velocity becomes perpendicular to initial velocity is. Thank your teacher & they could win INR75K.

Answer 2

Answer:

Hi myself Shrushtee. please follow the steps

Explanation:

The initial speed is υ

0

=10. The horizontal component of velocity is

υ0 cos600

=0.5υ0

=5

At the highest point of the trajectory of the projectile, its speed is 5m/s which is half the initial speed.

The time taken to reach the maximum height (when the vertical component of the speed becomes 0) is

0=υ0cos600=−gt

t= 10

10( 23 )

t=0.866sec

The required time is t=0.866 sec

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