a ball is thrown straight up from the edge of the roof of a building a second ball is dropped from the roof 1 s later if the height of the building is 20 m what must be the initial speed of the projection of first ball if both ball are to be hit at the same time
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A ball was thrown vertically upward from a 80m building. It reaches the ground after 10 seconds. What is the initial velocity if G=10m/s^2?
Let us take the top of the building as the origin of the coordinate system, upward direction as positive and downward direction as negative. A ball is thrown upwards with velocity u m/s from the top of the building ie the origin. It reaches the ground after 10s. We are required to find u, given that g= - 10 m/s² ( minus sign as g is directed downward).
Position of the ball after 10s = -80m ( minus sign as the ground is downwards from the origin).
Using the relation,
s = u t + ½ a t²
In this equation s= -80m, u=?, a = -10 m/s²; t=10s. Substituting in the equation we get,
-80 m = 10 u - ½ ×(10 m/s²)×100s²,
=> -80 m = 10 u - 500 m
=> 10 u = 420,; u = 42 m/s.
The ball was projected with a velocity of 42 m/s directed upwards.
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