HELP! Physics 1 Questions! Please!?

Question

An object initially at rest falls a distance h. use the formula y= - 1/2 gt^2 to express the time (t) it takes to fall in terms of h and g.

If the object is not initially at rest, but is moving horizontally with a speed v it will go a distance x-vt during the fall. use the time from part 1 to solve for v in terms of h, g, and x.

a balls is shot horizontally. its initially 1 m above ground. it hits the ground after traveling 2 m horizontally. what was the balls initial horizontal speed?

the ball is shot at an angle (theta) above horizontal. it is initially at ground level.. the distance it will travel before returning to the ground is x= v^2 sin (2 theta)/g. use the speed v of earlier ? and find the distance it will travel if (theta) = 30 degrees.

Given the range is x=v^2 sin (2theta)/g what angle should the ball be **** in order for it to travel the longest distance.

please show calculations! do any of the problems you can!

Answer 1

So you need help on all of these?? The first couple are quite easy. But I guess it would help to confirm what you think you understand -- so here goes.

**An object initially at rest falls a distance h. use the formula y= 1/2 gt^2 to express the time (t) it takes to fall in terms of h and g.

(I deleted the minus sign. At this point, it just confuses things.)

This is algebra.
h = (1/2)*g*t^2
t^2 = 2*h/g
t = sqrt(2*h/g)

**If the object is not initially at rest, but is moving horizontally with a speed v it will go a distance x-vt during the fall. use the time from part 1 to solve for v in terms of h, g, and x.

You meant to say x=vt.
If it went a distance x, v = x/t
and you know t from above.

**a balls is shot horizontally. its initially 1 m above ground. it hits the ground after traveling 2 m horizontally. what was the balls initial horizontal speed?

How long did it take to fall 1 m? Use y= 1/2 gt^2
That time, multiplied by the horizontal speed = the 2 meters.

**the ball is shot at an angle (theta) above horizontal. it is initially at ground level.. the distance it will travel before returning to the ground is x= v^2 sin (2 theta)/g. use the speed v of earlier ? and find the distance it will travel if (theta) = 30 degrees.

Just plug the velocity from the problem above into
x= v^2 sin (2 theta)/g
to find x.

**Given the range is x=v^2 sin (2theta)/g what angle should the ball be **** in order for it to travel the longest distance.

This means: if v and g are constant, what value of theta will give the max value of x when you evaluate the equation. If you think about how sine varies, you should get it.