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Motion in a Plane

Section: Physics

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Question : 61 of 94

Marks: +1, -0

A player kicks a football with an initial speed of

25\,\mathrm{ms}^{-1}

at an angle of 45° from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion ?

g=10\,\mathrm{ms}^{-2}

[27 Aug 2021 Shift 2]

$h_{\max}=10\,\mathrm{m}, T=2.5\,\mathrm{s}$
$h_{\max}=15.625\,\mathrm{m}, T=3.54\,\mathrm{s}$
$h_{\max}=15.625\,\mathrm{m}, T=1.77\,\mathrm{s}$
$h_{\max}=3.54\,\mathrm{m}, T=0.125\,\mathrm{s}$

Solution:

Given, initial speed,

u=25\,\mathrm{ms}^{-1}

Angle of projection,

\theta=45^{\circ}

Let, the maximum height

=H

Time taken to reach,

H=t

Acceleration due to gravity

g=10\,\mathrm{ms}^{-2}

Velocity at maximum height along

Y

-axis,

v_y=0\,\mathrm{ms}^{-1}

As we know that, for motion along

Y

-axis,

\because\; v_y^2-u_y^2=2gH

\therefore\; 0^2-(25\sin 45^{\circ})^2=-2\times 10\times H

\Rightarrow -312.5=-20H

\because\; H=15.625\,\mathrm{m}

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

\therefore\; =\sqrt{\frac{2H}{g}}=\sqrt{\frac{2\times 15.625}{10}}

\; =1.767=1.77\,\mathrm{s}

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