Projectile Motion-Shooting a Ball

[atbruick]

Homework Statement

A ball is shot from the top of a building with an initial velocity of 15 at an angle = 45 above the horizontal. What are the x and y components of the initial velocity? If a nearby building is the same height and 59 m away, how far below the top of the building will the ball strike the nearby building?

Homework Equations

Trig equations of sin and cos
sinTheta=opp/hyp cosTheta=adj/hyp

The Attempt at a Solution

I thought the x component at initial velocity would be zero, and the y component was sin45=y/15, which I got to be 11; Either one of these or both are wrong, my homework doesn't tell me if one is right or not. Not sure how to approach this because I thought the sin for y was correct.

 

[paco_uk]

If the x component of the initial velocity was zero then the ball would be traveling straight upwards. How have you calculated this?

 

[atbruick]

oh opps so just like you calculated y by using sine use cosine to get x. Thank you! I also just tried to find the second part, the distance it hits below the roof; I needed to find t so I used X=Xo+Vxot, 59=0+11t, and got t to be 5.36. Then plugged that into Y=Yo+Vyot-.5at^2, and got Y=0+11(5.36)-.5(-9.80)(5.36^2) and got 200 which was wrong. What am I plugging in wrong in the equation?

 

[paco_uk]

I think you are using all the right equations but you need to be careful with minus signs.

If you are regarding upwards as positive (a good idea) then your expression for y should read:

$$Y=Y_0+V_y(0)+\frac{1}{2} a t^2$$

Then you can consider in which direction acceleration due to gravity points.

Also watch out for rounding errors. You might find that you need more decimal places in your value for Vx and Vy.

 

[rwisz]

I would like to clarify that the x component of the initial velocity is not zero. The initial velocity of the ball has both horizontal and vertical components, which can be calculated using the trigonometric equations you have mentioned. The x component can be calculated as cos45 * 15 = 10.6 m/s, while the y component can be calculated as sin45 * 15 = 10.6 m/s.

To find the distance below the top of the nearby building where the ball will strike, we can use the equation for projectile motion:

y = y0 + v0y * t - 1/2 * g * t^2

Where:
y = vertical displacement
y0 = initial height (height of the building)
v0y = initial vertical velocity
g = acceleration due to gravity (-9.8 m/s^2)
t = time

We can set y = 0, since the ball will strike the nearby building at ground level. Substituting the values, we get:

0 = 45 + 10.6 * t - 1/2 * (-9.8) * t^2

Solving for t, we get t = 2.5 seconds.

Now, we can use the equation for horizontal displacement:

x = x0 + v0x * t

Where:
x = horizontal displacement
x0 = initial horizontal position (distance between the two buildings)
v0x = initial horizontal velocity
t = time

Substituting the values, we get:

x = 59 + 10.6 * 2.5

Solving for x, we get x = 84.5 meters.

Therefore, the ball will strike the nearby building 84.5 meters below the top of the building.