6: A body throws a ball vertically up. It returns to the ground after 5 seconds. Find the velocity with which the ball is thrown
up
1 answer:
Answer:
Explanation:
Ball rises and falls in equal periods. Because it goes by same speeds and same route. Therefore t_{rise} = t_{fall} =2.5 s
In free fall and vertical throw movement: x= (1/2).a.t^{2}
Here, we can find the height of the ball
x= (1/2).5.(2.5^{2}) = 15.625 m
From the attaches V-t graphics, x=V.t/2
15.624=V.(2.5)/2
V=12.5 m/s
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Answer:
900 m ( 0.9 km)
Explanation:
from the question we have the following
first person's height (h) = 100 m above the earths surface
distance to the horizon that can be seen (s) = 15 km = 15,000 m
second person's height (h1) = 36 m
distance to the horizon that can be seen (s1) = ?
take note that the distance (s) a person can see = square root of the person's height
s = x K
where k is a constant
from the height and the distance the first person can see we can get the value of K
15000 = x K
15000 = 10 x K
K = 150
now putting the value of K and the height of the second person into the equation we can get the distance into the horizon the person can see
s = x 150
s = 900 m
therefore the second person who is 36 m above the surface can see 900 m ( 0.9 km) into the horizon
Answer:
Explanation:
I was unable to submit this answer after typing it, so i create a document file that clearly shows and depicts all explanations for the question. OR
You can check the three picture slides i attached as well for the explanation.
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