How do you do this physics problem?

Question

I did it, but the answer doesn't seem to make much sense. Someone good at physics please just do it for me to make sure that I'm right.

A 0.2 kg rock is projected from the edge of the top of a building with an initial velocity of 9.21 m/s at an angle 43 degrees above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 14 m from the base of the building.

Assume: The ground is level and that the side of the building is vertical. The acceleration of gravity is 9.8 m/s^2

Part of the issue I think is that I don't know whether the acceleration should be positive or negative....I thought acceleration in the y direction was always negative, but then I was confused since the problem says it as a positive number

Answer 1

vx = v cos theta
vy = v sin theta

Using this convention, my vy will be positive and h will be positive (up), so I want negative acceleration (down).

y(t) = h + vy t - 1/2 gt^2 = 0 at landing

Solve the quadratic equation to get the landing time. Use the positive answer.

t = (- v sin theta +- sqrt ((v sin theta)^2 - 4(-g/2)(h)) )/ (2(g/2))

= -v sin theta / g + sqrt ((v sin theta / g)^2 + 2h/g)

horizontal distance = vx * t
= v cos theta
* (-v sin theta / g + sqrt ((v sin theta / g)^2 + 2h/g))

They give you v, h, and theta. You know g. Plugnchug to get the distance.

Answer 2

Whether the acceleration is positive or negative depends on which direction is chosen as positive. If you choose downward direction as positive then the acceleration of gravity is 9.8 m/s^2. If you choose upward direction as positive then the acceleration of gravity is -9.8 m/s^2. Since we USUALLY (not always) choose upward direction as positive so the acceleration of gravity is negative. In the question 9.8 m/s^2 is just the magnitude.
In horizontal direction: 14 = 9.21 cos (43) * t
In vetical direction (upward positive): -h = 9.21 sin (43) * t - 1/2 * 9.8 * t^2.

Answer 3

1) you don´t need the mass
The horizontal velocity is9.21*cos 43 =6.74m/s
14 = 6.74*t so the rock is in the aire2.08 s
Taking reference at the top of the building the equation of movement is
e = -4.9t^2+9.21 sin 43 *t = -4.9t^2+6.28 t
and
v= -9.8t +6.28
At maximum height v= 0 so t = 6.28/9.8=0.64s
e= 2.01m above the top of the building
From the maximum height to the ground the equation is
6.01 = +4.9 t^2 ( the rock moves downward and you must take g positive and initial velocity at maximum height is 0
t=1.11 s
Remember if up g<0 if down g>0