A body projected vertically upwards crosses a point at a height h twice in its journey at instants t1 and t2 second. The maximum height reached by the body is

Question

A body projected vertically upwards crosses a point at a height h twice in its journey at instants t1 and t2 second. The maximum height reached by the body is (A) (g/4)(t1 + t2)2 (B) g((t1 + t2/4))2 (C) 2g((t1 + t2/4))2 (D) (g/4)(t1 t2)

Tardigrade Admin · Accepted Answer

Time taken by the body to reach the point

A

is

t_{1}

(During upward journey).
Solution

The body crosses this point again (during the downward journey) after

t_{2}, i e,

the body takes the time

(t_{2} - t_{1})

to come again at the point

A .

So, the time taken by the body to reach at point

B

(a maximum height).

t = t_{1} + (\frac{t _{2} - t _{1}}{2})

[∵ T im e o f a sce n d in g = T im e o f d esce n d in g]

t = \frac{t _{1} + t _{2}}{2}

So, maximum height

H = \frac{1}{2} g t^{2}

= \frac{1}{2} g (\frac{t _{1} + t _{2}}{2})^{2}

= 2 g (\frac{t _{1} + t _{2}}{4})^{2}

Q. A body projected vertically upwards crosses a point at a height $h$ twice in its journey at instants $t_{1}$ and $t_{2}$ second. The maximum height reached by the body is

$\frac{g}{4} (t_{1} + t_{2})^{2}$

$g (\frac{t _{1} + t _{2}}{4})^{2}$

$2 g (\frac{t _{1} + t _{2}}{4})^{2}$

$\frac{g}{4} (t_{1} t_{2})$

Q. A body projected vertically upwards crosses a point at a height h twice in its journey at instants t1​ and t2​ second. The maximum height reached by the body is

4g​(t1​+t2​)2

g(4t1​+t2​​)2

2g(4t1​+t2​​)2

4g​(t1​t2​)

Q. A body projected vertically upwards crosses a point at a height $h$ twice in its journey at instants $t_{1}$ and $t_{2}$ second. The maximum height reached by the body is

$\frac{g}{4} (t_{1} + t_{2})^{2}$

$g (\frac{t _{1} + t _{2}}{4})^{2}$

$2 g (\frac{t _{1} + t _{2}}{4})^{2}$

$\frac{g}{4} (t_{1} t_{2})$