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Q.

A ball thrown up vertically returns to the thrower after 6 s. The position of the ball after 4 s is ____ m.


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Detailed Solution

A ball thrown up vertically returns to the thrower after 6 s. The position of the ball after 4 s is 4.9 m.
The equation of motion of an object forms a relation between displacement and velocity and acceleration and time. These equations are usually for the one-dimensional motion of an object.
Here, as the ball is thrown upwards and comes down, so the final velocity of the ball is zero. Also, the ball takes 6 s to go up and to come down. So, only to go up, it takes 3 seconds (62). Now we can calculate the position of the ball from the top after 1 s (as 3 + 1 = 4 s).
For that, we can use the formula,
⇒h = ut + 12 gt2….(1)
Here, initial velocity u = 0,
Time t = 1 s,
Gravitational acceleration g = 9.8 m/s2
By putting these values in equation (1),
⇒h = 0 × 1 + 12 × 9.8 × 12
⇒h = 4.9 × 1 ⇒h= 4.9 m
Hence, the ball will be at 4.9 m below the maximum height it reached.
 
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