Single Choice

From the top of a $$64$$ metres high tower, a stone is thrown upwards vertically with a velocity of $$48 \text{ m/s}$$. The greatest height (in metres) attained by the stone, assuming the value of gravitational acceleration $$g = 32 \text{ m/s}^2$$, is (Note: This question was asked in Maths subject in JEE Mains $$2015$$ online exam held on $$11$$ April $$2015$$)

A100
Correct Answer
B88
C128
D112

Solution

$$u = 48 \text{ m/s}, \: a = - 32 \text{ m/s}^2$$
At the highest point, $$v = 0$$
Since, $$ v^2 - u^2 = 2as$$,
$$\Rightarrow |s| = \left | \dfrac{0^2 - 48^2}{2 \times 32}\right | $$ m
$$\Rightarrow |s| = 36 \: m$$
So, total height attained by the stone from the ground $$=$$ $$ 36 + 64 = 100 $$ m

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