Motion Equation with constant speed (Time sensitive )

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Motion Equation with constant speed (Time sensitive!!)

Homework Statement



A stone thrown horizontally from a height of 6.02m hits the ground at a distance of 12.90m. The initial speed is 11.64 m/s. Calculate the speed of the stone as it hits the ground. Neglect air resistance.

Homework Equations



v^2-u^2=2as
v^2=2as+u^2

The Attempt at a Solution



u (initial velocity): 11.64 m/s
a (acceleration): 9.81 m/s^2
s (displacement): 12.9-6.02=6.58 m

v^2=2*9.81*6.58+11.64^2
v^2=264.59
v=16.27 m/s

And that is not right? Any hints or helping suggestions?

Thanks!
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
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Hi lanzjohn, welcome to PF.

Your displacement calculation is wrong. 12.9 m is the horizontal displacement, where as 6.02 is the vertical displacement.
The horizontal velocity remains constant.
The vertical velocity increases. The final vertical velocity can be found by v^2=2as+u^2 Here u = 0.

The velocity with which the stone hits the ground is given by

V^2 = Vy^2 + Vx^2
 
  • #3
collinsmark
Homework Helper
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Hello lanzjohn,

Welcome to Physics Forums!

There two ways solve this particular problem, kinematics and conservation of energy. But since you are approaching this from the kinematics method, I'll concentrate on that.

(Method I) Kinematics:

Homework Equations



v^2-u^2=2as
v^2=2as+u^2

The Attempt at a Solution



u (initial velocity): 11.64 m/s
a (acceleration): 9.81 m/s^2
s (displacement): 12.9-6.02=6.58 m

v^2=2*9.81*6.58+11.64^2
v^2=264.59
v=16.27 m/s

And that is not right? Any hints or helping suggestions?

Thanks!
You need work through this problem in two dimensions. Work with each dimension separately, when dealing with vectors. Later, if you need to sum the vectors and find the resulting magnitude (a magnitude such as speed), you can do that using the Pythagorean theorem.

The stone moves at a constant velocity along the x-axis. But it is accelerating along the y-axis. But you have scrambled up your dimensions. For example, you had,
.....v^2=2*9.81*6.58+11.64^2
The 9.81 [m/s2] is the acceleration in the y-direction. But the 11.64 [m/s] is the initial velocity in the x-direction. You can't mix up directions like that.

Solve for the final velocity in the y-direction and the x-direction separately. Then use the Pthagorean theorem to find the final speed.

(Method II) Conservation of Energy.

It turns out that this second method is much easier. But if you haven't studied conservation of energy problems yet, I won't spoil it for you! :tongue2:

[Edit: Looks like rl.bhat beat me to the punch again.]
 

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