Answer:
D
Explanation:
f = ma
2 x 12 = 24
answer could differ since it's rolling down a ramp. if an angle is given our approach differs.
Wee can use here kinematics
as we know that
![y = v*t + \frac{1}{2} at^2](https://tex.z-dn.net/?f=y%20%3D%20v%2At%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20at%5E2)
for shorter tree we know that
![y = 0 + \frac{1}{2}*9.8 * 2^2](https://tex.z-dn.net/?f=y%20%3D%200%20%2B%20%5Cfrac%7B1%7D%7B2%7D%2A9.8%20%2A%202%5E2)
![y = 19.6 meter](https://tex.z-dn.net/?f=y%20%3D%2019.6%20meter)
now since we know that other tree is twice high
So height of other tree is y = 39.2 m
now again by above equation
![y = v*t + \frac{1}{2} at^2](https://tex.z-dn.net/?f=y%20%3D%20v%2At%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20at%5E2)
![39.2 = 0 + \frac{1}{2}*9.8 * t^2](https://tex.z-dn.net/?f=39.2%20%3D%200%20%2B%20%5Cfrac%7B1%7D%7B2%7D%2A9.8%20%2A%20t%5E2)
![t = 2.83 s](https://tex.z-dn.net/?f=t%20%3D%202.83%20s)
so the time taken is 2.83 s
solution:
1.6 m/s = 96 m/min (in other words, 1.6 m/s x 60 s/min)
96 m/min x 8.3 min = 796.8 m
![s=ut +\frac{1}{2}at^2\\there is no accleration mentioned so,\\s= uv\\8.3\times60=498(s)\\510\times1.6=816(m)](https://tex.z-dn.net/?f=s%3Dut%20%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%3C%2Fp%3E%3Cp%3Ethere%20is%20no%20accleration%20mentioned%20so%2C%5C%5C%3C%2Fp%3E%3Cp%3Es%3D%20uv%5C%5C%3C%2Fp%3E%3Cp%3E8.3%5Ctimes60%3D498%28s%29%5C%5C%3C%2Fp%3E%3Cp%3E510%5Ctimes1.6%3D816%28m%29)
Answer:
The answer to your question is : 521.8 m
Explanation:
Data:
Different heights
Time first object (tfo) = 10.7 s
Time second object (tso)= 14.8 s
Initial speed of both objects(vo) = 0 m/s
a = 9.81 m/s²
Formula:
h = vot + 1/2 (a)(t)² but vo = 0 so, h = 1/2 (a)(t)²
Then, height fo h = 1/2 (9.81)(10.7)² = 561.6 m
height so h = 1/2(9,81)(14.8)² = 1074.4 m
Difference in their heights = 1074.4 m - 561.6 m = 521.8 m