Question #284883

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?

___ s


1
Expert's answer
2022-01-05T04:21:22-0500

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 g=9.8ms2\small g = 9.8 ms^{-2} .
  • Then you can use s=ut+12at2\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) )

+30.9m=+7.95ms1×t+12×(+9.8)×t20=4.9t2+7.95t30.9[a quadratic function]t={1.83s3.45s[neglected]\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 1.83s\small 1.83\,s

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS