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Answer to Question #29258 in Mechanics | Relativity for slvi
Question #29258
a ball is projected upwards from the top of a tower with a velocity of 50m/s making an angle of 30 degree with the horizontal. the hieght of the tower is 70m. after how many seconds from the instant of throwing will the ball reach the ground?
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This is the formula expressing the distance in case of accelerated motion. In most general form it looks like this s=s0+vt+at^2/2 In our case we assume that h is the height of the ball and h=0 on the ground. So h increases in vertical direction. Also we have acceleration due to gravity, which acts downwards, that's why we take a=-g We take the term vt with "+" sign because a ball is projected upwards due to the statement. We have h0 in the formula, because initial position of the ball due to vertical axis is not zero: a ball is projected upwards from the top of a tower, the hieght of the tower is 70m. Hopefully, now it's clear.
from where did we get h=h0+vt-gt^2/2 ?
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