Asked by Jim
A ball is thrown upwards with an initial velocity of 16 m/s. What is the velocity of the ball 1.5 s later?
After the ball leaves the hand of the thrower, what is the value of acceleration acting on the ball as it is rising upwards?
9.8x1.5 = 14.7. 16-14.7=1.3 m/s after 1.5s.
The value of acceleration acting on the ball is -9.8 m/s, due to gravity.
Is this correct?
After the ball leaves the hand of the thrower, what is the value of acceleration acting on the ball as it is rising upwards?
9.8x1.5 = 14.7. 16-14.7=1.3 m/s after 1.5s.
The value of acceleration acting on the ball is -9.8 m/s, due to gravity.
Is this correct?
Answers
Answered by
Henry
V = Vo + g*t = 16 - 9.8*1.5 = 1.3 m/s.
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