Question #184267

A ball is thrown vertically upwards from a platform 8 feet above the ground. After 2 seconds, the

ball is 2 feet below the platform. Find the initial velocity of the ball.


Expert's answer

y=h+vt0.5gt22=8+2v0.5(32.174)(2)2v=29.2ftsy=h+vt-0.5gt^2\\2=8+2v-0.5(32.174)(2)^2\\v=29.2\frac{ft}{s}


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