ANSWERS: 2
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this one is simple; a) the time it is released is 0seconds... so when t=0 h(t) = h(0) = 2.1 (im assuming meters) b) to find the maximum hight you will have to take the derivative: h'(t) = -14.6t +8.25 CORRECTION! and now find when h'(t)=0; max height happens at t = 0.565s h(0.565) = 4.425m It is the same procedure for ever question of this type... (PS. i think this should be in Calculus category)
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There is not enough data to answer the second question. You need to know the initial speed of the ball.
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