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1) a flowerpot is dropped from the balcony of an apartment, 28.5m above the ground. at a time of 1.00s after the pot is dropped, a ball is thrown vertically downward from the balcony one storey below, 26.0m above the ground. the initial velocity of the ball is 12.0m/s[down]. does the ball pass the flowerpot before... 顯示更多 1) a flowerpot is dropped from the balcony of an apartment, 28.5m above the ground. at a time of 1.00s after the pot is dropped, a ball is thrown vertically downward from the balcony one storey below, 26.0m above the ground. the initial velocity of the ball is 12.0m/s[down]. does the ball pass the flowerpot before striking the ground? if so, how far above the fround are the two objects when the ball passes the flowerpot? plz show all steps 更新: 仲有無其他formula可以計到架？...
最佳解答:
hey man, i am here to make some change something wrong with the answers above this is because the ball was not thrown from the same positon as the flowerpot so if we assume the ball will pass the flowerpox in the air, the condition for that is not df =db where df refer to the displacement of the flowerpot from time=0 (sec) to time=t (sec) db refer to the displacement of the ball time = 1 (s) to time = t (s) it should be df=db+2.5m it is because the distance travel by the ball should be shorter than the flowerpot 2.5m the equation becomes 1/2(g)(t^2) = u(t-1)+1/2[(g)(t-1)^2]+2.5 after expanding some terms 1/2(g)(t^2) = ut-u+1/2(g)(t^2)-gt+g/2+2.5 0=ut-u-gt+g/2+2.5 after some rearrangement t=[u-(g/2)-2.5]/(u-g) notice that t is refer to the time travelled by the flowerpot to the position them will meet put u = 12.0m/s and g =9.8ms^-2 we have t=23/11s then df =1/2(g)(t^2)=1/2(9.8)(23/11)^2=(21.422)m (to 3 d.p.) since distance travelled by the flowerpot is only 21.422 m which is smaller than 28.5m the ball will pass the flowerpot at a height of (28.5-21.422)=7.08m ( to 3 sig. fig.)
其他解答:
First let g = 9.8m/s. then the flowerpot travels a distance d in time t: (i) df = (gt^2)/2. 1.00 second later the ball is thrown with initial velocity 12.0m/s and travels a distance d in time t-1 : (ii) db = 12(t-1) + [g(t-1)^2]/2 For the ball to pass the flowerpot they have to travel a same distance where df = db (gt^2)/2 = 12(t-1) + [g(t-1)^2]/2 (gt^2)/2 = 12t - 12 +(gt^2)/2 - gt - g/2 0 = 12t -gt -12 +g/2 (put in 9.8 for g) we have 0 = 12t -(9.8)t -12 + 9.8/2 0 = 2.2t - 7.1 t = 3.23s put t = 3.23 into (i) we have df = 51.1m since the height of the flowerpot's balcony is only 28.5m above ground, therefore, the ball will not pass the flowerpot before they hit the ground.F5B24A77BB847046
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發問:1) a flowerpot is dropped from the balcony of an apartment, 28.5m above the ground. at a time of 1.00s after the pot is dropped, a ball is thrown vertically downward from the balcony one storey below, 26.0m above the ground. the initial velocity of the ball is 12.0m/s[down]. does the ball pass the flowerpot before... 顯示更多 1) a flowerpot is dropped from the balcony of an apartment, 28.5m above the ground. at a time of 1.00s after the pot is dropped, a ball is thrown vertically downward from the balcony one storey below, 26.0m above the ground. the initial velocity of the ball is 12.0m/s[down]. does the ball pass the flowerpot before striking the ground? if so, how far above the fround are the two objects when the ball passes the flowerpot? plz show all steps 更新: 仲有無其他formula可以計到架？...
最佳解答:
hey man, i am here to make some change something wrong with the answers above this is because the ball was not thrown from the same positon as the flowerpot so if we assume the ball will pass the flowerpox in the air, the condition for that is not df =db where df refer to the displacement of the flowerpot from time=0 (sec) to time=t (sec) db refer to the displacement of the ball time = 1 (s) to time = t (s) it should be df=db+2.5m it is because the distance travel by the ball should be shorter than the flowerpot 2.5m the equation becomes 1/2(g)(t^2) = u(t-1)+1/2[(g)(t-1)^2]+2.5 after expanding some terms 1/2(g)(t^2) = ut-u+1/2(g)(t^2)-gt+g/2+2.5 0=ut-u-gt+g/2+2.5 after some rearrangement t=[u-(g/2)-2.5]/(u-g) notice that t is refer to the time travelled by the flowerpot to the position them will meet put u = 12.0m/s and g =9.8ms^-2 we have t=23/11s then df =1/2(g)(t^2)=1/2(9.8)(23/11)^2=(21.422)m (to 3 d.p.) since distance travelled by the flowerpot is only 21.422 m which is smaller than 28.5m the ball will pass the flowerpot at a height of (28.5-21.422)=7.08m ( to 3 sig. fig.)
其他解答:
First let g = 9.8m/s. then the flowerpot travels a distance d in time t: (i) df = (gt^2)/2. 1.00 second later the ball is thrown with initial velocity 12.0m/s and travels a distance d in time t-1 : (ii) db = 12(t-1) + [g(t-1)^2]/2 For the ball to pass the flowerpot they have to travel a same distance where df = db (gt^2)/2 = 12(t-1) + [g(t-1)^2]/2 (gt^2)/2 = 12t - 12 +(gt^2)/2 - gt - g/2 0 = 12t -gt -12 +g/2 (put in 9.8 for g) we have 0 = 12t -(9.8)t -12 + 9.8/2 0 = 2.2t - 7.1 t = 3.23s put t = 3.23 into (i) we have df = 51.1m since the height of the flowerpot's balcony is only 28.5m above ground, therefore, the ball will not pass the flowerpot before they hit the ground.F5B24A77BB847046
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