A projectile is fired horizontally with a velocity of $98 m / s$ from the top of a hill $490 m$ high. Find the velocity with which the projectile hits the ground. $(g = 9.8 m / s^{2})$

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Solution

Given : $u_{x} = 98 m / s$ $a_{y} = g = 9.8 m / s^{2}$ $u_{y} = 0$
Also time of flight $T = 10$ s (solved earlier)

x direction : $a_{x} = 0$ $⟹ v_{x} = u_{x} = 98 m / s$
y direction : $v_{y} = u_{y} + a_{y} T$ where $T = 10$ s
$⟹ v_{y} = 0 + 9.8 (10) = 98 m / s$
$∴$ Velocity with which the particle hits the ground $\to V = 98^i + 98^j$
$⟹$ $| \to V | = \sqrt{98^{2} + 98^{2}} = 98 \sqrt{2}$ m/s

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A projectile is fired horizontally with a velocity of 98m/s from the top of a hill 490m high. Find the velocity with which the projectile hits the ground. (g=9.8m/s2)

Given : ux=98m/s ay=g=9.8m/s2 uy=0Also time of flight T=10 s (solved earlier)x direction : ax=0 ⟹vx=ux=98m/sy direction : vy=uy+ayT where T=10 s ⟹vy=0+9.8(10)=98m/s∴ Velocity with which the particle hits the ground →V=98^i+98^j⟹ |→V|=√982+982=98√2 m/s

A projectile is fired horizontally with a velocity of $98 m / s$ from the top of a hill $490 m$ high. Find the velocity with which the projectile hits the ground. $(g = 9.8 m / s^{2})$