Projectile Motion gravity and acceleration

[2ndperiod]

Homework Statement

A ball is thrown horizontally from the top of a building 54.5 m high. The ball strikes the ground at a point 61 m from the base of the building.

The acceleration of gravity is 9.8 m/s^2 (9.8 meters per second squared)

1. Find the time the ball is in motion. (in units of s)
2. Find the initial velocity of the ball (in units of m/s)
3. Find the x component of its velocity just before it strikes the ground. (in units of m/s)
4. Find the y component of its velocity just before it strikes the ground.

HELP! please!

Homework Equations

Delta X = VX T

R = (Vsquared sin (2 x angle)) / g

Delta Y = (1/2) (Vyi + Vyf) (t)

Vyf = Vyi - (g)(t)

Delta Y = (Vyi)(t) - (1/2)(g)(Tsquared)

Vyfsquared = Vyisquared - (2)(g)(Delta Y)

The Attempt at a Solution

If someone could just tell me which formulas to use, that would be great. I know I need to solve for t for #1, and need to solve for Vyi for #2.

 

[tatiana]

First if you were to draw a visual of this problem you would get a triangle that would be easier for you to se ethat you need to find the value of the hypotenuse which is the velocity of the time in teh air before it gets 61m from the building. Then you can use that to sole for t in the equation.

 

[2ndperiod]

First if you were to draw a visual of this problem you would get a triangle that would be easier for you to se ethat you need to find the value of the hypotenuse which is the velocity of the time in teh air before it gets 61m from the building. Then you can use that to sole for t in the equation.

I did draw a diagram.

So my hypotenuse is V? And then what formula should I use to solve for t?

 

[fss]

Giving formulas is not very educational, and is basically doing the work for you. Here are some thinking points.

#1. The ball is thrown horizontally, that is in the +x direction. Since it is thrown horizontally, what is the initial velocity in the vertical direction? What does that tell you about the time it takes the ball to reach the ground, given that the x and y velocities are independent of each other?

#2. Since you know the time from #1, this is a very simple formula. You are given a distance (61 m from the base of the building) and you have calculated the time it takes to hit the ground. You should be able to calculate a velocity from that.

That should get you started, at least.

 

[rwisz]

For #3 and #4, I am not sure which formulas to use.



Hello, based on the information given, we can use the following equations to solve for the given parameters:

1. To find the time the ball is in motion, we can use the formula: Delta X = VX * T. We know that Delta X is 61 m and VX is the initial horizontal velocity (which we will solve for in #2). Therefore, we can rearrange the equation to solve for T: T = Delta X / VX. Plugging in the values, we get T = 61 m / VX.

2. To find the initial velocity of the ball, we can use the formula: R = (Vsquared * sin(2 * angle)) / g. In this case, the angle is 0 degrees since the ball is thrown horizontally. We know that R is the range, which is given as 61 m, and g is the acceleration due to gravity, which is 9.8 m/s^2. Therefore, we can rearrange the equation to solve for V: V = sqrt(R * g / sin(2 * angle)). Plugging in the values, we get V = sqrt(61 m * 9.8 m/s^2 / sin(0)). This simplifies to V = sqrt(61 m * 9.8 m/s^2 / 0), which is undefined. This means that the ball was not thrown with an initial velocity, but rather dropped from the building. Therefore, the initial velocity is 0 m/s.

3. To find the x component of the velocity just before it strikes the ground, we can use the formula: Delta X = VX * T. We know that Delta X is 61 m and T is the time we solved for in #1, which is 61 m / VX. Therefore, we can rearrange the equation to solve for VX: VX = Delta X / T. Plugging in the values, we get VX = 61 m / (61 m / VX). This simplifies to VX = 61 m / 1, which is 61 m/s.

4. To find the y component of the velocity just before it strikes the ground, we can use the formula: Vyf = Vyi - (g)(t). We know that Vyi is the initial vertical velocity, which we can assume to be 0 m/s since the