Question #ffc23

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Question #ffc23

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Answer 1 · 2016-05-29T16:44:08

$\frac{5}{4} s and 7.8125 m$ $5/4s and 7.8125m$

Explanation:

Let the balls meet t sec after the second ball is thrown.
So the 1st ball will then complete (t+1) sec of its journey.

The imitial velocity of the 1st ball is zero .So it will fall a height $h = \frac{1}{2} \cdot g \cdot {(t + 1)}^{2}$ $h=1/2*g*(t+1)^2$

The 2nd ball's initial velocity being $30 \frac{m}{s}$ $30m/s$ it should fall the same height during t sec
So $h = 30 \cdot t + \frac{1}{2} \cdot g \cdot t^{2}$ $h=30*t+1/2*g*t^2$

Inserting $g = 10 m s^{- 2}$ $g=10ms^-2$ and equating two height we can write
$\frac{1}{2} \cdot 10 \cdot {(t + 1)}^{2} = 30 t + \frac{1}{2} \cdot 10 \cdot t^{2}$ $1/2*10*(t+1)^2=30t+1/2*10*t^2$
$\Rightarrow 5 {(t + 1)}^{2} - 5 t^{2} = 30 t$ $=>5(t+1)^2-5t^2=30t$
$\Rightarrow 5 (2 t + 1) = 30 t$ $=>5(2t+1)=30t$
$\Rightarrow 2 t + 1 = 6 t$ $=>2t+1=6t$
$\Rightarrow t = \frac{1}{4} s$ $=>t=1/4s$

So the ball will meet $(t + 1) = \frac{1}{4} + 1 = \frac{5}{4} s$ $(t+1)=1/4+1=5/4s$ after the 1st ball starts.

During this time it will fall
$h = \frac{1}{2} \cdot 10 \cdot {(\frac{5}{4})}^{2} = \frac{125}{16} m = 7.8125 m$ $h=1/2*10*(5/4)^2=125/16m=7.8125m$

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Question #ffc23

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