Two students are on a balcony 21.4 m above the street. One student throws a ball

Question

Two students are on a balcony 21.4 m above the street. One student throws a ball vertically downward at 13.3 m/s; at the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down.

a) What is the velocity of the ?rst ball as it strikes the ground?
b) What is the velocity of the second ball as it strikes the ground?
c) What is the di?erence in the time the balls spend in the air?
d) How far apart are the balls 0.611 seconds after they are thrown?

Explanation / Answer

v^2=13.3^2+2*10*21.4 v=24.569 m/s 2. v=24.569m/sec 3.t1=24.569-13.3/10=1.1269 t2=24.569+13.3/10=3.7869 diff in time =2.66 sec 4.s1=13.3*0.611+5*.611*.611=9.99m s2=-13.3*0.611+5*.611*.611=-6.25m thus distance between them=15.24 m i have assumed value of g=10m/s^2 if in ur question it is 9.8 there will b slight...

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