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A projectile is fired from point 0 at the edge of a cliff, with initial velocity components of V(ox)= 60 m/s and V(oy)= 175 m/s. The projectile rises and then falls into the sea at point P. The time of flight of the projectile is 40.0 s, and it experiences no appreciable air resistance in flight. What is the magnitude of the velocity of the projectile 21.0 s after it is fired?

Respuesta :

Answer:

The magnitude of velocity=60.82 m/s

Explanation:

Given that

V(ox)=60 m/s

V(oy)=175 m/s

We know that in projectile motion

In y-direction

The time of flight T=40 s

Time to reach at maximum height is 20 sec and that point component of velocity in y-direction will become zero.

At 20 sec object will have zero velocity in y-direction.

So at t=21 sec

Velocity V= g Δt

V= 10 (21-20) m/s

V=10 m/s

It means that velocity will be in negative y-direction(downward).

In x-direction

The x-component of velocity will remain constant in x-direction

Vx=60 m/s

The magnitude of velocity after 21 s

[tex]Magnitude=\sqrt{10^2+60^2} [/tex]

The magnitude of velocity=60.82 m/s

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