As you stand by the side of the road, a car approaches you at a constant speed, sounding its horn, and you hear a frequency of 8
1 answer:
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Answer:
The second ball
Explanation:
Both balls are under the effect of gravity, accelerating with exactly the same value. The first ball is dropped, therefore its initial velocity is zero. Since the second ball has horizontal and vertical velocity components, its initial velocity is given by:
The vertical component is zero, however, it has a horizontal velocity, so its initial speed is not zero, therefore the secong ball has the greater speed at ground level.
Answer:
Data:-vi=om/s (b/c as in question penny is dropped from building means before coming to ground its initial state or velocity was considered as zero ) now distance or height h=380m and now we have to find the final velocity vf=? and the time t=?
Explanation:
So applying second eq of motion s=vit+1/2×gt² (here we have taken a gravity b/c when ever body is in vertical position then acceleration due to gravity is applied ) s=0×t+1/2×gt² , s=0+1/2×9.8×t² ,380=4.9t² we have to find t so 4.9t²=380 , t²=380÷4.9 , t²=77.55 now sq root on b/s
so t=8.806s and now apply 1st eq o²f motion to find out vf so vf=vi+gt , vf=0+9.8×8.806 ,vf=86.298 and if you want to verify that either this is answer is correct or not so put the value of t in second eq of motion and if you got distance same as give in the question so your value of t is considered as correct likewise s=vit+1/2gt² , s=0+1/2×9.8(8.806)²,s=4.9×77.55 ,s=380m (proved) I hope it would be helpfull