A person throws a rock straight up into the air. At the moment it leaves the per

Question

A person throws a rock straight up into the air. At the moment it leaves the person's hand it is going 35 mph. When the rock reaches its peak, how fast is it going and what is the magnitude and direction of its acceleration?

Speed
mph

Acceleration
m/s2

Right Down East West Up Left South North
A person throws a rock straight up into the air. At the moment it leaves the person's hand it is going 35 mph. When the rock reaches its peak, how fast is it going and what is the magnitude and direction of its acceleration?

Speed
mph

Acceleration
m/s2

Right Down East West Up Left South North

Explanation / Answer

Given that the initial speed of rock is u = 35 mph *--------------------------------------------------------------- At its peak height (Maximum height) final speed of the rock is zero That is at peak speed v = 0 mph The acceleration of the body does not change during the motion, it is constant and it is equal to the acceleration due to gravity g = 9.8 m/s^2 The acceleration due to grvity always points down ward direction.

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