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Rashid [163]
3 years ago
6

A ball is thrown vertically upward with a speed of 25.1 m/s. How high does it rise? The acceleration due to gravity is 9.8 m/s2

. Answer in units of m.
How long does it take to reach its highest point? Answer in units of s.
How long does it take the ball to hit the ground after it reaches its highest point? Answer in units of s.
What is its velocity when it returns to the level from which it started? Answer in units of m/s.
Physics
1 answer:
malfutka [58]3 years ago
3 0
 <span>Things you should know before answering... At its absolute highest point the velocity of a ball will be 0 m/s. 
1. You have Vi,Vf, and a. You're looking for d so the best equation is: vf^2=Vi^2+2ad (0=26.5^2+2(-9.81)d). 
0=702.5-19.62d 
-702.5=-19.62d 
d=35.793 meters 

2. You now have Vi,Vf,a, and d. You are looking for t. There are many equations you can use but to keep things simple we will use: Vf=Vi+at (0=26.5-9.81t) 
-26.5=-9.81t 
t=2.701 seconds 

3. You just have to double the time it took to reach the top. 

4. Depends on what the question means. Theoretically at the starting point if the ball is no longer travelling it's at 0. If not then you should know that the descent on something falling vertically is mirrored exactly to the ascent meaning that the final speed is 26.5 m/s. 

Vf^2=Vi^2+2ad 
Vf^2=0+2(-9.81)35.793 
vf=sqrt(702.259) 
vf=26.5001</span><span>
</span>
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B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}

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\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k

\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})

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B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}

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