Answer to Question #284883 in Mechanics | Relativity for happy

Question

A ball is thrown directly downward with an initial speed of 7.95 m/s, from a height of 30.9 m. After what time interval does it strike the ground?

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Answer 1

Explanations & Calculations

• The ball falls under gravity which is almost constant for the low heights above earth.
• Hence you can use any of the 4 motion equations suitably to find the unknown value.
• You are given the initial velocity, height at which the ball was released, and asked for the time it takes to reach the ground.
• The hidden quantity/ directly not given but you need to know is the gravitational acceleration $\small g = 9.8 ms^{-2}$ .
• Then you can use $\small s = ut+\frac{1}{2}at^2$ downwards from the starting point to find the time. (then the downward direction is +, all the vector quantities direct downward are taken + while any others upwards are taken negative) )

$\qquad\qquad \begin{aligned} \small +30.9\,m&=\small +7.95\,ms^{-1}\times t+\frac{1}{2}\times(+9.8)\times t^2\\ \small 0 &=\small 4.9t^2+7.95t-30.9\cdots[\text{a quadratic function}]\\ \small t&=\small \begin{cases} \small 1.83\,s\\ \small -3.45\,s\cdots[\text{neglected}] \end{cases} \end{aligned}$

• Then the time it would take is $\small 1.83\,s$