the eaquation of projectile is y=16x-5x^2/4 .the horizontal range is
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Answered by
168
Thats very simple!
Do you know that x=0 and x= R, y=0? Do you get this, why is it so? Because the projectile is at the surface of projection at x=0 while at x=R also the projectile comes to the surface of projection after flight, hence y=0.
So, we just have to find the roots of y= 16x-5x²/4= 0
So,
16x -5x²/4 = 0
x(16 -5x/4) = 0
✓ x=0 and
✓ 16-5x/4=0
ie x = 64/5 = 12.8 units
Hence, horizontal range = 12.8 units ✓
Hope you understand the concept.
All the best!
Do you know that x=0 and x= R, y=0? Do you get this, why is it so? Because the projectile is at the surface of projection at x=0 while at x=R also the projectile comes to the surface of projection after flight, hence y=0.
So, we just have to find the roots of y= 16x-5x²/4= 0
So,
16x -5x²/4 = 0
x(16 -5x/4) = 0
✓ x=0 and
✓ 16-5x/4=0
ie x = 64/5 = 12.8 units
Hence, horizontal range = 12.8 units ✓
Hope you understand the concept.
All the best!
Answered by
22
Answer:
16x-5x^2/4=0
16x =5x^2/4
x =64/5
x =12.8
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