Question

A projectile is thrown with a velocity of
10√2 ms-1 at an angle of 45° with the
horizontal. The time interval between the
moments when the speeds are 125 ms-1 is
(8=10ms-2)​

Answer 1

Answer:

1 sec

Step-by-step explanation:

Need to find : The time interval between the moments when the speeds are √125 m/s

projectile is thrown with a velocity of 10√2 m/s

angle of 45°

Horizontal Velocity = 10√2Cos45  = 10 m/s

Vertical Velocity = 10√2Sin45  = 10 m/s

Horizontal Velocity will remain constant = 10

Let say at time t the Velocity = √125 m/s

=> Vertical Velocity² + Horizontal Velocity² =( √125 )²

=> Vertical Velocity²  + 10² = 125

=> Vertical Velocity² = 25

=> Vertical Velocity = ±5

Applying V = U + at

5 = 10 + (-10)t

=> t = 0.5 sec

-5 = 10  + (-10)t

=> t = 1.5 sec

√125 m/s velocity would be at t =0.5 sec & 1.5 sec

Time interval between = 1.5  - 0.5 = 1 sec