Question

From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.40 m/s and angle of 15.0° below the horizontal. It strikes the ground 5.00 s later. (a) If the base of the building is taken to be the origin of the coordinates, with upward the positive y-direction, what are the initial coordinates of the ball? (Use the following as necessary: y0. Assume SI units. Do not substitute numerical values; use variables only.)

Answer 1

Answer:

$y_{o}=v_{o}*sin(\alpha )*t + 1/2*gt^{2}$

Explanation:

Kinematics equation, with gravity as deceleration

$y =y_{o}-v_{o}*sin(\alpha )*t - 1/2*gt^{2}$

If the base of the building is taken to be the origin of the coordinates, then :

$y_{o}$ is the initial coordinates of the ball

for t=5.00, the ball strikes the ground ⇒ y=0m

then:

$y_{o}=v_{o}*sin(\alpha )*t + 1/2*gt^{2}$

Answer 2

Answer:

We want to find the value of y0, so let's analyze the problem:

We only should care for the y-axis movement, so this may be a one dimencional problem.

First, the acceleration in the y-axis is the gravitational one, so we have:

a(t) = -g

For the velocity we integrate over time:

v(t) = -g*t + v0

and we have a constant of integration v0, that is the initial velocity in the y-axis, in this case we have v0 = 8.4m/s*sin(15°) where the initial velocity is positive, so the ball is initial throwed upwards.

At last, the position in Y is

Y(t) = -(g/2)*t^2 + v0*t + y0

where y0 is the initial position.

Now we know that in t = t0 = 5s, the ball reaches the ground, so we have:

Y(t0) = 0 = (-g/2)*(t0)^2 + v0*t0 + y0

then we can find the initial height y0 by isolating it in the last equation:

y0 = (g/2)*(t0)^2 -v0*t0

And it is represented as variables only, as wanted.