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Question

(B) Let the first ball be thrown from the ground at $y=0$ with initial velocity $v$. Its height at time $t=0.8 \,s$ is given by $h = vt - \frac{1}{2}g

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) Let the first ball be thrown from the ground at $y=0$ with initial velocity $v$. Its height at time $t=0.8 \,s$ is given by $h = vt - \frac{1}{2}gt^2$.Substituting the values: $h = v(0.8) - \frac{1}{2}(10)(0.8)^2 = 0.8v - 3.2$.The second ball is dropped from a height of $20 \,m$. Its height from the ground at time $t=0.8 \,s$ is given by $y = H - \frac{1}{2}gt^2$.Substituting the values: $y = 20 - \frac{1}{2}(10)(0.8)^2 = 20 - 3.2 = 16.8 \,m$.Since the balls are at the same height, $h = y$.Therefore, $0.8v - 3.2 = 16.8$.$0.8v = 20$.$v = \frac{20}{0.8} = 25 \,ms^{-1}$.

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