Question

Two stones are thrown towards each other simultaneously, one vertically downwards from the top of the
tower of height 360 m with velocity 10 m/s and other from the ground vertically upwards with velocity
50 m/s. Find the time and the distance from the ground at which they collide. [Take,g=10 m/s2]
plzz reply fast​

Answer 1

height of the tower is 360 let the distance covered by top stone is x then, distance covered by bottom stone is 360-x

for top,

h=ut+(1/2)gt^2

x=10*t+5t^2------(1)

5t^2=x-10t--------(2)

for bottom,

360-x=50*t-5t^2

by euq. (2),

360-x=50t-x+10t

360=60t

t=6 sec

putting t=6 sec in eq(1),

x=10*6+5*6*6

x=60+180

x=240m

and

360-x=360-240=120m

so the time taken to meet is 6 sec

and distance from top is 240m and from bottom is 120m

Answer 2

Answer:

height of the tower is 360 let the distance covered by top stone is x then, distance covered by bottom stone is 360-x

for top,

h=ut+(1/2)gt^2

x=10*t+5t^2------(1)

5t^2=x-10t--------(2)

for bottom,

360-x=50*t-5t^2

by euq. (2),

360-x=50t-x+10t

360=60t

t=6 sec

putting t=6 sec in eq(1),

x=10*6+5*6*6

x=60+180

x=240m

and

360-x=360-240=120m

so the time taken to meet is 6 sec

and distance from top is 240m and from bottom is 120m

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