Physics help... emergency thank you?

Question

An boy throws a stone of mass 1kg. The stone is released with an initial velocity 5m/s
at an angle 30o above horizontal while it is at the height 1.28m. How far from the boy
will the stone land? What is the speed of the stone when it hits the ground? What is
the highest altitude the stone reaches?

Answer 1

The horizontal component of the stone's velocity is V0*cosø. It maintains that velocity as long as it is in the air. Call that time ta, so the distance from the boy is V0*ta*cosø. You need to find ta. This time is composed of two parts: ascent (t1) and descent (t2). The ascent time is the time it takes for the vertical component to reach zero under gravitational deceleration of g = 9.8 m//sec^2. The initial vertical velocity is V0*sinø, and the time of ascent t1 is given by V0*sinø = g*t1, or

t1 = (V0*sinø)/g.

The altitude is h = V0*t1 - 0.5*g*t1^2 + 1.28

EDIT: The above applies to vertical component only, so should be h = V0*t1*sinø - 0.5*g*t1^2 + 1.28


The descent time t2 is given by h = 0.5*g*t2^2; find t2 using h from above.

Now you can get the horizontal distance, since ta = t1 + t2.

The velocity when it hits the ground is acceleration times descent time, or g*t2

You now have enough to answer all of the questions.

Answer 2

The mass of the stone is irrelevant to basic projectile problems.

Start with the horizontal and vertical components of the initial velocity: v(h)=5cos30=4.33 m/s and v(v)=5sin30=2.50 m/s.

An initial vertical velocity of 2.50 m/s will become zero in t=v/a=2.5/9.8=0.255 seconds. Use the distance formula to get d=1/2at^2=4.9*(0.255)^2=0.32 meters for the maximum height above its initial elevation, making the height above the ground 1.6 meters.

Use the distance formula again to solve for t=sqrt(2d/a)=sqrt((2*1.6)/9.8)=sqrt(0.3265)=0.57 seconds to hit the ground from this height.

The final velocity is v=at=9.8*0.57=5.6 m/s and it lands (0.255+0.57)*4.33=3.6 meters from the boy.

Answer 3

that is been too long when you consider that I labored issues such as those, so i do no longer remember the math. yet, on the 2d difficulty, draw a sketch. the respond ought to be an exceptionally small quantity. remember, 10mm=1cm. 10cm=1dm. 10dm=1m. all of the forces pronounced are to the left of the pivot component and the load of the stick (rod?) is to the right of the pivot. pondering the place the forces are utilized, and their instructions, the consequence ought to be very small.