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A ball is thrown vertically upward. It has a speed of 10 m/sec when it has reached one half of its maximum height. How high does the ball rise? Take $g = 10$ m/s$^2$ -
A
5 m
B
15 m
C
10 m
D
20 m
Detailed Solution
Let the maximum height be $h$. At the top the speed is zero.
Consider the motion from the half-way point (speed 10 m/s) to the top, a further height of $\dfrac{h}{2}$.
Using $v^2 = u^2 - 2gs$: $0 = (10)^2 - 2 \times 10 \times \dfrac{h}{2}$
$10^2 = 2(g)\left(\dfrac{h}{2}\right) = 10h$
$h = \dfrac{100}{10} = 10$ m
So the ball rises to a maximum height of 10 m.
Consider the motion from the half-way point (speed 10 m/s) to the top, a further height of $\dfrac{h}{2}$.
Using $v^2 = u^2 - 2gs$: $0 = (10)^2 - 2 \times 10 \times \dfrac{h}{2}$
$10^2 = 2(g)\left(\dfrac{h}{2}\right) = 10h$
$h = \dfrac{100}{10} = 10$ m
So the ball rises to a maximum height of 10 m.