Question

A body is projected horizontally with velocity 10 m/s from the top of
tower of height 20 m, on reaching the ground its displacement is
(g=10 m\s2)



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Answer 1

• Since the object returned to origin point , i.e , from where it was launched
• Thus the displacement of the object would be = 0

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Answer 2

$s=20\sqrt{2}\ m$ is the displacement of the body at the end of the projectile.

Explanation:

Given:

• horizontal velocity of projection of the projectile, $u_x=10\ m.s^{-1}$

• height of the tower from where the projectile is projected, $h=20\ m$

• acceleration due to gravity, $g=10\ m.s^{-2}$

• Now as the projectile is given only initial horizontal velocity of projection therefore its vertical velocity of projection is zero, $u_y=0\ m.s^{-1}$

Time of flight:

(the time taken by the body to fall on the ground)

using eq. of motion,

$h=u_y.t+\frac{1}{2} a.t^2$

$20=0+0.5\times 10\times t^2$

$t=2\ s$

Now the horizontal distance traveled by the body in this time:

$s_x=v_x.t$

$s_x=10\times 2$

$s_x=20\ m$ from the base of the tower.

Now its displacement form the initial point can be given by the Pythagoras theorem applied to the triangle formed by height of the building, distance of the body from the base of the tower and the displacement between the final point and the initial point of projection of projectile.

$s=\sqrt{(20)^2+(20)^2}$

$s=20\sqrt{2}\ m$

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TOPIC: projectile motion

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