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
Answers
Answered by
0
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.
Answered by
1
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
Please mark me as brainleist
Similar questions