You throw a stone straight upward with initial speed 15 m/s. If you want to throw a second stone straight upward with speed 20 m/s and have it collide with the first stone at the exact instant the first stone reaches the peak of its motion, when should you throw up the second stone?
In order to answer this question, let's first analyze the motion of the stone with the initial speed of 15 m/s, the first stone.
The time it will take the stone to reach the peak (the highest point) of its motion is determined by the equation
`v_f = v_0-g*t`
` ` , where `v_f = 0` ` ` because at the highest point the velocity of the ball is zero, and g is the gravitational acceleration, 9.8 m/s^2.
The time it takes the first stone to reach the peak is `t=15/9.8 = 1.53 s` ` `
The maximum height of the first stone can be found from this equation of motion:
`v_f^2 - v_0^2 = -2gh`
From here, the height is `h = (v_0^2)/(2g) = 15^2/(2*9.8) = 11.5 m` .
The second stone has to hit the first stone at the moment when the first stone is at its maximum height. This means, the second stone's motion has to be such that at t = 1.53 seconds after the first stone was thrown, the second's stone's height has to be 11.5 meters.
Let's find how long it will take the second stone, with initial velocity 20 m/s, to reach 11.5 meters:
`h = v_0t - g*t^2/2`
`11.5 = 20*t - 9.8*t^2/2`
`4.9t^2 - 20t + 11.5 = 0`
Use the quadratic formula to solve this quadratic equation:
`t = (20+-sqrt(20^2 - 4*4.9*11.5))/(2*4.9)`
`t = 3.39 ` s and `t = 0.69` s.
The second stone will reach the height of 11.5 meters twice: at t = 0.69 s, on its way up, and at t = 3.39 s, on its way down.
Since it take the first stone 1.53 second to reach the same height, there are two ways to throw the second stone, so it will hit the first stone when it is at this height:
the time 1.53 - 0.69 = 0.84 seconds AFTER the first stone was thrown
the time 3.39 - 1.53 = 1.86 seconds BEFORE the first stone was thrown.
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