Grade 11Mechanics
a boy can throw a stone up to a height 10 m the maximum horizontal distance that boy can throw same stone is
3 Answers
Puja Ivaturi
Given vertical height = 10m. if we took it as max vertical height
then, u2/2g = 10m => u2/g = 20
As horizontal projectile range = u2sin2∂/g
So, range is maximum when ∂=45 degrees
Hence, maximum horizontal range is u2/2g = 20,m
Rameshwar Puri
Vertical height attend by ball
We know at maximum height final velocity become zero
V=0
H = u^2 / (2g) ……[1]
For the same body with same initial initial velocity the horizontal distance travelled when thrown at an angle (θ) is given as
R = u^2 Sin(2θ) / g
Maximum horizontal distance is given as
R = u^2 / g ……[2]
Compare equations 1 and 2
R = 2 H
R = 2 × 10
R = 20 m
Rishi Sharma
Dear Student,
Please find below the solution to your problem.
Given vertical height = 10m.
If we took it as max vertical height then,
u^2/2g = 10m
=> u2/g = 20
As horizontal projectile range = u^2sin2∂/g
So, range is maximum when∂=45 degrees
Hence, maximum horizontal range is
u^2/2g = 20,m
Thanks and Regards
Please find below the solution to your problem.
Given vertical height = 10m.
If we took it as max vertical height then,
u^2/2g = 10m
=> u2/g = 20
As horizontal projectile range = u^2sin2∂/g
So, range is maximum when∂=45 degrees
Hence, maximum horizontal range is
u^2/2g = 20,m
Thanks and Regards
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