Solve for e. T= e^2/9
1 answer:
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Answer:
38.6 m/s
Step-by-step explanation:
The motion of the ball is a projectile motion, which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction
Therefore we have to analyze the horizontal and vertical motion separately.
Along the horizontal direction, the velocity is constant during the motion, since there are no forces acting in this direction. So the horizontal velocity 3 seconds after the launch will be the same as the velocity at the launch:
The vertical velocity instead changes according to the suvat equation:
where
is the initial vertical velocity
is the acceleration due to gravity
t is the time
Therefore, after t = 3 s,
So the velocity after 3 seconds is < 25, -29.4 > m/s. The magnitude of the velocity is