A ball is thrown upward, reaches its highest point, and then comes straight down

Question

A ball is thrown upward, reaches its highest point, and then comes straight down. Air resistance is negligible and can be ignored. At the tip pity top of its path its velocity is momentarily minimal. a) The acceleration at the top is 9.8 m/s^2 down. b) The acceleration at the top is 0 m/s^2 c) The acceleration at the top is 9.8 m/s^2 up d) Since the ball is in free fall, its acceleration is undetermined. e) None of the above is correct. A ball is tossed in the air. It moves upward, reaches its highest point and falls back downward. a) Sketch a velocity-time and an acceleration-time graph for the ball from the moment it leaves the thrower's hand until the moment just before it reaches her hand again. Consider the positive direction to be upward. b) Circle on the velocity vs time graph the point where the ball has reached its highest point.

Explanation / Answer

15 whenever we throw an object straight up the acceleration direction is always downwards since the force due to gravity is pulling the object so this is the reason the objects gets slower as its moves upwards and at the apex it stops but acceleration is still there only its final velocity is zero since a=dv/dt , now the object comes down with the same acceleration and its speed increases . the acceleration magnitude will remain the same i.e 9.8m/s2 and its always downwards thats the reason while in upward motion we take -g since its direction in downward and in downward motion we take it +g in formulas.

so the answer will be option (a) direction is always downward with 9.8m/s2

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