A stone projectile hitting the target

[rudransh verma]

Homework Statement

A stone is thrown by aiming directly at the center P of the pic hanging on wall. The stone leaves the starting point horizontally with a speed of 6.75m/s and strikes the target at point Q 5cm below point P. Find the horizontal distance between the starting point of stone and the target.

Relevant Equations

$$(x-x0)=ut$$

$$(y-y0)=ut-1/2gt^2$$
$$y= -1/2gt^2$$
$$t^2=-1/98$$
If I get t I will be able to solve for x=ut

 

[berkeman]

Correct, so you only need to find out how long it takes an object to drop 5cm. Since the stone is thrown horizontally, it is the same as just dropping it.

Nice LaTeX!

 

[berkeman]

BTW, your last equations should be more like:

$$\Delta y = \frac{1}{2}g t^2$$
$$t^2 = \frac{2 \times 0.05}{9.8}$$

Since the motion and acceleration are in the same direction...

(EDIT -- Fixed error 0.005 --> 0.05 to represent 5cm)

 

[rudransh verma]

BTW, your last equations should be more like:

$$\Delta y = \frac{1}{2}g t^2$$
$$t^2 = \frac{2 \times 0.05}{9.8}$$

Since the motion and acceleration are in the same direction...

Can I say the value given in the question of displacement is negative 5 cm.

 

[berkeman]

Can I say the value given in the question of displacement is negative 5 cm.

Yes you can, but it's important to also then say that the acceleration of gravity "g" is pointing in the negative direction. Otherwise you end up with the square root of a negative number, which would give you an imaginary result (and would be wrong in this case).

$$t^2=-1/98$$

 

[rudransh verma]

Since the stone is thrown horizontally, it is the same as just dropping it.

I guess you are trying to say that the stone will cover the horizontal and vertical distance in same time t

 

[berkeman]

I guess you are trying to say that the stone will cover the horizontal and vertical distance in same time t

Yes. That is how many projectile motion problems are solved. Often, we use the constant horizontal velocity of the projectile and the known horizontal distance to solve for the time t, and then plug that back into the vertical motion equation (which has the quadratic term because of the constant downward acceleration of gravity) to find the vertical position.

In your current problem, things are a bit different, but we are still solving for the time t first to then use it in a 2nd equation.