A stone is dropped from a height of 300m and at the same another stone is projected vertically upwards with a velocity of 100ms-1.Find when and where the two stones meet?

Question

Yashwant · Answer

Given the height with Wich the ball is left freely is 300m Let the two bodies will meet at a distance `x` meters from the heighest pont , let they meet in time `t` seconds X=ur+1/2gt^2X=5t^2

Sheharyaar · Answer

for this question we can use relative speed concept total distance d = 300m relative speed = (100 × 0) m/s [since the objects are moving in the opp direction] therefore time t = d/s = 300/100 = 3s so by s = ut + ½ gt^2 = 0(3) + ½ * 10 * 3^2 = 5 * 9 =45 m I hope you like my answer☺

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A stone is dropped from a height of 300m and at the same another stone is projected vertically upwards with a velocity of 100ms-1.Find when and where the two stones meet?

A stone is dropped from a height of 300m and at the same another stone is projected vertically upwards with a velocity of 100ms-1.Find when and where the two stones meet?

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A stone is dropped from a height of 300m and at the same another stone is projected vertically upwards with a velocity of 100ms-1.Find when and where the two stones meet?

A stone is dropped from a height of 300m and at the same another stone is projected vertically upwards with a velocity of 100ms-1.Find when and where the two stones meet?

2 Answers

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