A rock is shot vertically upward from the edge of the top of a tallbuilding. The

Question

A rock is shot vertically upward from the edge of the top of a tallbuilding. The rock reaches its maximum height above the top of thebuilding 1.16 s after being shot. Then,after barely missing the edge of the building as it falls downward,the rock strikes the ground 5.50 s afterit was launched. (a) With what upward velocity was the rockshot?
1 m/s

(b) What maximum height above the top of the building is reached bythe rock?
2 m

(c) How tall is the building?
3 m (a) With what upward velocity was the rockshot?
1 m/s

(b) What maximum height above the top of the building is reached bythe rock?
2 m

(c) How tall is the building?
3 m

Explanation / Answer

v=u+at v = final speed, u = initial speed t=1.16s a= -9.81m/s2 0= u + (-9.81)(1.16) u= 11.3796 m/s ----------------------------------------------------------------- v2=u2+2ay y= distance travelled 0=11.37962+2(-9.81)y y=6.60m ------------------------------- y(building)=ut+0.5at2, where t=5.5s = 11.3796(5.5)+.5(-9.81)(5.5)2 = -85.78845 m the negative sign indicate it is below the startingpoint. the height of building is 85.78845 m. = -85.78845 m the negative sign indicate it is below the startingpoint. the height of building is 85.78845 m. hope this helps

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