A watermelon cannon fires a watermelon vertically up into the air at a velocity of +9.5m/s, starting from an initial position of
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<h2>The magnitude 24 (
) of the acceleration of the particle when the particle is not moving.</h2>
Explanation:
Given,
A particle moving along the x-axis has a position given by
m ........ (1)
To find, the magnitude ( ) of the acceleration of the particle when the particle is not moving = ?
Differentiating equation (1) w.r.t, 't', we get
⇒ ....... (2)
⇒
⇒
⇒ t = 2 s
Again, differentiating equation (2) w.r.t, 't', we get
Put t = 2, we get
Thus, the magnitude 24 ( ) of the acceleration of the particle when the particle is not moving.