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Grade 11Mechanics

A hiker stands on the edge of a cliff 490 metre above the ground and throws the stone horizontally with an initial speed of 15 metre per second.The speed with which it hits the ground is...?

Profile image of Jwalin Raval
9 Years agoGrade 11
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1 Answer

Profile image of Vikas TU
9 Years ago
in horizontal direcn. net accln. = 0
for horzntl. dircn. therfore,
 
Vx  = u + 0 = 15 m/s  …....................(1)
 
For vertical direcn.,
490 = 0 + 0.5gt^2
calculate t from here,
then from first eqn.
Vy = 0 + gt
and get Vy.
 
Resulatant speed = root(Vx^2 + Vy^2)