Projectile motion, initial height and range given, find initial velocity

[dainceptionman_02]

Homework Statement

At what initial speed would a projectile have to be ejected at initial angle 35° to the horizontal, initial height 3.3km, in order to land at a range of x = 9.40km?

Relevant Equations

y = (tanθ.)x - (gx^2)/[2(vcosθ.)^2]

the answer to this question uses the above formula with the tangent function and solves for the initial velocity,
i used the equation (v.sinθ^2) = (v.sinθ)^2 - 2gΔy, setting final velocity equal to zero and solving for initial velocity. this kinematic equation gives a different answer. can someone tell me why the formula i used did not work?

 

[Philip Koeck]

Homework Statement: At what initial speed would a projectile have to be ejected at initial angle 35° to the horizontal, initial height 3.3km, in order to land at a range of x = 9.40km?
Relevant Equations: y = (tanθ.)x - (gx^2)/[2(vcosθ.)^2]

the answer to this question uses the above formula with the tangent function and solves for the initial velocity,
i used the equation (v.sinθ^2) = (v.sinθ)^2 - 2gΔy, setting final velocity equal to zero and solving for initial velocity. this kinematic equation gives a different answer. can someone tell me why the formula i used did not work?

Two remarks:

The final velocity is not zero. Why would it be?

Don't start with an equation. Start by thinking about what's happening. Think of the x and the y-direction separately.

 

[dainceptionman_02]

Two remarks:

The final velocity is not zero. Why would it be?

Don't start with an equation. Start by thinking about what's happening. Think of the x and the y-direction separa

the final velocity is zero in the x and y-direction for this particular situation. shouldn't that equation work?

 

[Philip Koeck]

the final velocity is zero in the x and y-direction for this particular situation. shouldn't that equation work?

In the moment the projectile hits the ground its velocity is not zero.
If it gets stuck in the ground it becomes zero, but that's a completely different story.

 

[dainceptionman_02]

In the moment the projectile hits the ground its velocity is not zero.
If it gets stuck in the ground it becomes zero, but that's a completely different story.

lol, well it does get stuck in the ground. it is a volcano launching a rock that "lands" at the base of the mountain.

 

[PeroK]

lol, well it does get stuck in the ground. it is a volcano launching a rock that "lands" at the base of the mountain.

The impact with the ground is not covered by projectile motion. The equations of projectile motion describe the motion before impact with the ground. The final velocity in this case is the velocity of the projectile at the instant of impact. Which is definitely non-zero.

 

[haruspex]

The impact with the ground is not covered by projectile motion.

To elaborate.. the SUVAT equations, including that listed as Relevant Equation, assume constant acceleration. Once the rock touches the ground the acceleration changes and the equation no longer applies. You have to pick the instant before that.

 

[Gargi]

Homework Statement: At what initial speed would a projectile have to be ejected at initial angle 35° to the horizontal, initial height 3.3km, in order to land at a range of x = 9.40km?
Relevant Equations: y = (tanθ.)x - (gx^2)/[2(vcosθ.)^2]

the answer to this question uses the above formula with the tangent function and solves for the initial velocity,
i used the equation (v.sinθ^2) = (v.sinθ)^2 - 2gΔy, setting final velocity equal to zero and solving for initial velocity. this kinematic equation gives a different answer. can someone tell me why the formula i used did not work?

You can first use R = ucosθ*t and substitute the value of R and θ. You will get somewhere around 11.75 as the product of u*t
Now, you can use the equation S = usinθt - 1/2gt^2 and substitute the values of S (-3.3, as per the sign convention), θ and u*t. This will give you the time of flight(t) which you can substitute in the product of u*t.

 

[haruspex]

You can first use R = ucosθ*t and substitute the value of R and θ. You will get somewhere around 11.75 as the product of u*t
Now, you can use the equation S = usinθt - 1/2gt^2 and substitute the values of S (-3.3, as per the sign convention), θ and u*t. This will give you the time of flight(t) which you can substitute in the product of u*t.

The thread is a bit old.
Thank you for trying to help the OP, but it is important to focus on the question being asked by the OP, which can be different from the task they were given.
In the present case, why didn't the method they used work? That was answered in posts #4, #6 and #7.