PLEASE HELP! I don't get it?

Question

A 0.15-kg ball is thrown so that it leaves your hand 2.0 m above the ground and has a speed of 25 m/s at an angle of 45 degrees upward from the horizontal. What is its speed just before it hits the ground? How high does the ball rise?

(assume there is no friction)

Answer 1

actually im going to have to agree completely with yanke4life8787 on this one. You actually gave a wrong answer for the maximum height you divided by g not 2g therefore your answer is quite wrong. You are correct in saying that you do not need the time. However solving the problem by obtaining the time is still a great way to get the answer and can quite often come in handy for example if another part of the question was to find the duration of the flight than two answers would have been answered in one shot. Also Yanke4life8787 stated he was solving for Velocity which he probally thought was what the asker intended to as and if that was not then he now has a better understanding that he would of by your answer. I think it was a waste of your time to repeat exacttly what yanke4life8787 answered none the less giving an incorrect answer for the height.

Answer 2

The mass doesn't matter, and you don't need to know the time of flight.....

Vyi = 25sin45° = 17.678 m/s

Vx = 25cos45° = 17.678 m/s (constant throughout flight)

Vyf² -Vyi² = 2gh → Vyf = √(2gh + Vyi²) = √(2*9.8*2.0+17.678²)

Vyf = 18.754 m/s

Vf = √(Vx²+Vyf²) = 25.7725 m/s

Since you only asked for speed, not velocity, the answer is complete here.

Rise = Vi²*sin²45°/g = 25²*sin²45°/9.8 = 31.888 m

Hmax = Rise + 2.0 = 33.888 m

Answer 3

The mass of the ball is completely irrevelent. However what is relevant would be the initial height initial speed and direction.

Vf^2=Vo^2+2a(delta_y)

Vf=0 at the top

Vo=25sin(45)  (the sin function is used to describe the y direction and the cos is used for the x direction

a=-9.8m/s^2

delta_y=(-Vo^2/2a)   =(25sin45)^2/(2*-9.8)=15.94meters

however not that delta y = 15.94 meters since you started 2m above ground the true height where the ground =o is 17.94 m

to find the velocity before it hits the ground you must know how much time the trip takes to do this use delta_y=Vosin(45)T+(1/2)aT^2  set the quadratic = to zero after making delta y -2 since going from 2 m to 0  final - initial=-2 get the time then plug into the x any y equations

VFx=25cos(45)t

VFy=25sin(45)t

VF=sqrt(VFy^2+VFx^2)

now in order to be complete you must give a direction

theta=arctan(VFy/VFx)

the direction you get should make sense so you may have to add 180 degrees to your answer depending on what it gives you

if y is positive then you may only be in quadrent 1 or 2

if y is negative then you may onbe be in quad 3 or 4

if x if positibe than you may only be in quad 1 or 4

if x is negative than you may onle be in quad 2 or 3

from that figure out what your true angle is than you will have a complete answer.

Goodluck!