Mechanics -projectile motion calculating angle

[umm...]

Homework Statement

A stone is thrown from a building of height 40m and reaches a maximum height of 49m from the ground. The stone strikes the ground at a distance of 10m from the foot of the building. find the angle of elevation from which the stone is projected. Assuming the ground level is horizontal.

Homework Equations

Hmax=Vo²sin²θ/2g

The Attempt at a Solution

from the hmax formula i found Vosinθ=13.28m/s

 

[azizlwl]

Homework Statement

A stone is thrown from a building of height 40m and reaches a maximum height of 49m from the ground. The stone strikes the ground at a distance of 10m from the foot of the building. find the angle of elevation from which the stone is projected. Assuming the ground level is horizontal.

Homework Equations

Hmax=Vo²sin²θ/2g

The Attempt at a Solution

from the hmax formula i found Vosinθ=13.28m/s

y=u_yt- 0.5gt²
x=u_xt

 

[Infinitum]

Hmax=Vo²sin2θ/g

That is incorrect. Hmax=h + Vo²sin²θ/2g is correct. The equation you've written gives the range of the projectile(not the maximum height.) To the OP,
What is the horizontal distance traveled by the stone in terms of velocity and t? And the vertical distance traveled by the stone in terms of t?

 

[umm...]

from y=uyt- 0.5gt2
t=4.8s
if substituted in x=uxt
Ux=2.08s=Vo
hence sinθ=6.37 which is invalid

 

[azizlwl]

That is incorrect. Hmax=h + Vo²sin²θ/2g is correct. The equation you've written gives the range of the projectile(not the maximum height.)

Sorry about that, assuming 'H' for horizontal and and I was wrong assuming vertical displacement=0, which not in this case.

 

[azizlwl]

from y=uyt- 0.5gt2
t=4.8s
if substituted in x=uxt
Ux=2.08s=Vo
hence sinθ=6.37 which is invalid

How do you find t when there are 2 unknown variables, t and u_y. in 1 equation?

 

[umm...]

How do you find t when there are 2 unknown variables, t and u_y. in 1 equation?

Uy is Usinθ=13.28m/s
which i found from Hmax equation:
Usinθ=√9×2×9.8=13.28

 

[azizlwl]

The equation Hmax=Vo2sin2θ/2g applies to when vertical displacement equal to zero.
In this case the stone does not drop to the same level as it was projected.

 

[umm...]

The equation Hmax=Vo2sin2θ/2g applies to when vertical displacement equal to zero.
In this case the stone does not drop to the same level as it was projected.

yeah and that is why is did not use height 49 but i used height 9m...hope I'm understanding what you meant :S

 

[azizlwl]

Ok it is correct if you substitute 9m for the equation.
What you get is the time taken from start to the same level(on descending) of the flight.
But there another 40m to travel down. Now you are making 2 segments of flight.
It is best to remember simple formula for displacement of constant acceleration.

Δs=s₀+ut ±.5at²

If at t=0 and s<sub>0[/SUB=0
Δs=ut ±.5at²

If at t=0 , s0=0 and a=0
Δs=ut

From these 2 equations you can derive all other equations depends on Δs.</sub>

 

[rl.bhat]

Use -y = Usinθ*t - 1/2*g*t^2. Substitute the values for y, Usinθ and g and solve for t.

 

[umm...]

from y=uyt- 0.5gt2
t=4.8s
if substituted in x=uxt
Ux=2.08s=Vo
hence sinθ=6.37 which is invalid

Use -y = Usinθ*t - 1/2*g*t^2. Substitute the values for y, Usinθ and g and solve for t.

here this is how i get...did i substitute wrong?

 

[Kearnsy101]

I am new to this thing and don't know how to post my own question so that's the first thing. Second my problem is I know displacement is 20m, acceleration is -5.4m/s, final velocity is 0m/s, coefficient of friction=0.55, the object slowing down weighs 1000kg, can i find initial velocity and the time it took to slow down and if so how?

 

[rl.bhat]

-40 = 13.28*t - 4.9*t^2
Now solve for t.

 

[azizlwl]

from y=uyt- 0.5gt2
t=4.8s
if substituted in x=uxt
Ux=2.08s=Vo
hence sinθ=6.37 which is invalid

t=4.8s
UCosθt=10
UCosθ=10/4.8

USinθ/UCosθ=13.28 x 4.9/10
θ=81°

 

[umm...]

t=4.8s
UCosθt=10
UCosθ=10/4.8

USinθ/UCosθ=13.28 x 4.9/10
θ=81°

thanks :D