You throw a rock straight up at a speed of 12m/s. With what speed must a friend on the roof of a building that is 39m above you release point throw a second rock so that the two rocks collide 1/6 of the way to the top (i.e. 6.5m above you release point)?
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We have to assume both rocks are released at the same t = 0. And the height of a projectile at any time t is y(t) = h + Uyt - 1/2 gt^2. h is the height the projectile is launched from with vertical speed Uy.
So we solve 6.5 = Uy t - 4.9 t^2 and 4.9 t^2 - 12t + 6.5 so that t = 1.64, which is how long the rock climbs to reach 6.5 m. We assume h = 0 as nothing was given re that.
Then 6.5 = 39 + Uy 1.64 - 4.9*1.64^2 and Uy = (4.9*1.64^2 - 35 + 6.5)/1.64 = -9.34 mps downward. ANS. The minus sign for Uy means the toss is downward
So we solve 6.5 = Uy t - 4.9 t^2 and 4.9 t^2 - 12t + 6.5 so that t = 1.64, which is how long the rock climbs to reach 6.5 m. We assume h = 0 as nothing was given re that.
Then 6.5 = 39 + Uy 1.64 - 4.9*1.64^2 and Uy = (4.9*1.64^2 - 35 + 6.5)/1.64 = -9.34 mps downward. ANS. The minus sign for Uy means the toss is downward
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taking g=10m/s
the distance would be 1.2245m not 6.5m
get the value of t from the equation
5t square - 12t + 1.2245 = 0
and the get the value of u (initial vel. of throwing) from
5t square + ut - 37.78 =0
the value of u should be your answer acc. to me ...
the distance would be 1.2245m not 6.5m
get the value of t from the equation
5t square - 12t + 1.2245 = 0
and the get the value of u (initial vel. of throwing) from
5t square + ut - 37.78 =0
the value of u should be your answer acc. to me ...