From the top of the tower 45 m high, two stones are released. One vertically downwards and the other with a horizontal velocity of 30 m/s. How long will each stone to strike the ground and how far from the tower will each stone strike the ground
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Answered by
2
U=30m/s
V=0
S=45m
We know
2as=v^2-u^2
A=(0)-(30×30)/2×45
A=-10m/s
We also know that
A=v-u/t
-10×t= 0-30
T = -30/-10
T=3sec
V=0
S=45m
We know
2as=v^2-u^2
A=(0)-(30×30)/2×45
A=-10m/s
We also know that
A=v-u/t
-10×t= 0-30
T = -30/-10
T=3sec
Answered by
0
The stone that is thrown horizontally, so there is no vertical component of the velocity. Hence both stone will strike the ground in same time. Now vertically both ball is starting from rest with acceleration g=10 m/s2 so and H=45m so we have
t=2Hg−−−√=2×4510−−−−√=3s.
So bot stone will strike the ground after 3 second.
Now the stone dropped vertically will touch the ground in the same spot that is horizontal distance will be zero. The stone that is thrown will continue to move with constant horizontal speed of 30 m/s, since there is no component of force and hence acceleration in horizontal direction. So the stone will strike in 30×3=90m from the tower.
t=2Hg−−−√=2×4510−−−−√=3s.
So bot stone will strike the ground after 3 second.
Now the stone dropped vertically will touch the ground in the same spot that is horizontal distance will be zero. The stone that is thrown will continue to move with constant horizontal speed of 30 m/s, since there is no component of force and hence acceleration in horizontal direction. So the stone will strike in 30×3=90m from the tower.
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