Question

A body is projected with a velocity of 40m/s at an angle of 45 degre with horizontal surface what will be the range of projection

Answer 1

R=(u^2sin2A)/g
given u=40m/s
A=45
g=10
R=(40X40)(sin2(45))/10
=1600(sin90)/10
=1600/10=160m

Answer 2

Given that,

Initial velocity of a body, v = 40 m/s

Angle of projection, $\theta=45^{\circ}$

To find,

Horizontal range of projection.

Solution,

The formula for the range of projection of a projectile is given by :

$R=\dfrac{u^2\sin 2\theta}{g}$

g is acceleration due to gravity

$R=\dfrac{(40)^2\sin 2(45)}{9.8}\\\\R=163.26\ m$

So, the range of projection is 163.26 meters.

Learn more,

Projectile motion

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