Asked by Nifes
A stone projected vertically upwards with a velocity of 20m/s. Two seconds later, a second stone is similarly projected with the same velocity. When the two stones meet, the second one is rising at a velocity of 10m/s. Neglecting air resistance, calculate the (i) lenght of time the second stone is in motion before they meet. (ii) velocity of the first stone when they meet (take g=10m/s^2).
Answers
Answered by
Henry
1st stone: V = Vo + g*Tr.
0 = 20 - 10Tr, Tr = 2 s. = Rise time.
1. 2nd stone: V = Vo + g*t.
10 = 20 - 10t, t = 1 s. in motion.
2. 1st stone: V = Vo + g*t. = 0 + 10*1 = 10 m/s. When they meet.
0 = 20 - 10Tr, Tr = 2 s. = Rise time.
1. 2nd stone: V = Vo + g*t.
10 = 20 - 10t, t = 1 s. in motion.
2. 1st stone: V = Vo + g*t. = 0 + 10*1 = 10 m/s. When they meet.
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