Question

1.
A particle is projected with speed 10 m/s at an angle 60° with horizontal. Find :
(a) time of flight
(b) range
(c) maximum height
(d) velocity of particle after one second.
(e) velocity when height of the particle is 1 m​

Answer 1

Answer:

The initial speed is υ0=10. The horizontal component of velocity is

υ0cos600=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=1010(23)

t=0.866sec

The required time is t=0.866 sec

Answer 2

Answer:

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