Projectile Motion gravity and acceleration

AI Thread Summary
A ball is thrown horizontally from a height of 54.5 m and lands 61 m away, with gravity at 9.8 m/s². To find the time of flight, the vertical motion can be analyzed using the formula for free fall, while the horizontal distance allows for calculation of initial velocity. The initial vertical velocity is zero since the ball is thrown horizontally, meaning the time to hit the ground is determined solely by the height. Once the time is known, the horizontal velocity can be calculated using the distance traveled. Understanding the independence of x and y motion is crucial for solving the problem effectively.
2ndperiod
Messages
4
Reaction score
0

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.
 
Physics news on Phys.org
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.
 
tatiana said:
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?
 
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.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top