Question

Velocity of a stone projected , 2 second before it reaches the maximum height makes an angle of 53° with horizontal then velocity at the highest point will be

Answer 1

Answer:

Answers : (1)

Acc to ques, 2 seconds before it reaches the maximum height it makes an angle of 53 deg

=> Initial velocity of y-component will be (Uy)=Usin53

=> Initial velocity of x-component will be (Ux)=Ucos53

Therefore, Vy=Uy-gt

=>0=Usin53-20

=>U=25m/s

Now, At highest point the y-component velocity is 0, only x-comp velocity exists which remians constant throughout the projectile motion.

Hence, At highest point the overall velocity of stone will be equal to the x-component velocity

ie, Ucos53=25x3/5=15m/s ANSWER

Answer 2

Answer:

Acc to ques, 2 seconds before it reaches the maximum height it makes an angle of 53 deg

=> Initial velocity of y-component will be (Uy)=Usin53

=> Initial velocity of x-component will be (Ux)=Ucos53

Therefore, Vy=Uy-gt

              =>0=Usin53-20

              =>U=25m/s

Now, At highest point the y-component velocity is 0, only x-comp velocity exists which remians constant throughout the projectile motion.

Hence, At highest point the overall velocity of stone will be equal to the x-component velocity

ie, Ucos53=25x3/5=15m/s ANSWER