a student can throw a ball vertically to a maximum height of 40 m .the same student can throw the ball in horizontal direction to a maximum distance of ?
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Heya.......!!!
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According to the question , a student can throw a ball vertically to a maximum height 40 m.
We have to take out the maximum distance in horizontal direction .
Firstly the vertical height which is 40 m is given by the formulae :
=> u^2 / 2g = 40
We know that Horizontal range is given by the formulae :
=> u^2 sin2θ / g
For maximum range θ = 45°
putting θ = 45° , we get sin2θ = 1
that is => u^2 / g
Now if see carefully the formulae of vertical height i.e = u^2 / 2g = 40
from this we can take out the value of u^2 /g
→ u^2 / g = 80 m
♦ Maximum horizontal range = 80 m
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Hope It Helps You ☺
___________________________
According to the question , a student can throw a ball vertically to a maximum height 40 m.
We have to take out the maximum distance in horizontal direction .
Firstly the vertical height which is 40 m is given by the formulae :
=> u^2 / 2g = 40
We know that Horizontal range is given by the formulae :
=> u^2 sin2θ / g
For maximum range θ = 45°
putting θ = 45° , we get sin2θ = 1
that is => u^2 / g
Now if see carefully the formulae of vertical height i.e = u^2 / 2g = 40
from this we can take out the value of u^2 /g
→ u^2 / g = 80 m
♦ Maximum horizontal range = 80 m
=================================
Hope It Helps You ☺
lovely03:
hi...
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hope it helps u sir ❤❤❤❤❤❤❤❤❤
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