Question

when does R=4H and R=2H? related with projectile motion.​

Answer 1

Answer:

R=

g

16sin2θ

h=

2g

16sin

2

θ

R

h

=

g

16sin

2

θ

×

16sinθ

g

=

4

1

tanθ

=tanθ=

R

4h

X=

R

2

+(4h)

2

sinθ=

R

2

+16h

2

4h

R

max

=

16h

2

2h(R

2

+16h

2

)

R

max

=

8h

R

2

+2h

Answer 2

Answer:

Apply (1) to find velocity components:

On the x-axis: ux=u0cos(θ) (since there is no acceleration on x-axis)

On the y_axis: uy=u0sin(θ)−gt (sine ay=−g )

Apply (3) on the y axis, note that when the object reaches maximum height, vy=0

=> 02−(u0sin(θ))2=2(−g)H

=> H=(u0sin(θ))22g

Now, calculate the total flight time of the object, at t = 0 uy=u0sin(θ) , when the object touches the ground, uy=−u0sin(θ)

Apply (1): −u0sin(θ)=u0sin(θ)−gt1

=> t1=2u0sin(θ)g

To find the horizontal range:

R=ux.t1=u0cos(θ)2u0sin(θ)g

which finally gives, R=u20g2sin(θ)cos(θ)

So the relation between H and R is

R=4Hcos(θ)sin(θ)

Using the same method (set of 3 equations), we can solve a harder problem from this: find the range and maximum height when throw an object at a height h, and see the difference.l