Question

A ball A is thrown vertically upward from the top of a 28-m-high building with an initial velocity of 4 m/s. At the same instant another ball B is thrown upward from the ground with an initial velocity of 20 m/s. Part A Determine the height from the ground at which they pass. Express your answer to three significant figures and include the appropriate units hValue Units Submit Request Answer Part B Determine the time at which they pass Express your answer to three significant figures and include the appropriate units

Answer 1

Answer:

A) 19.994 m B) 1.750 seconds

Explanation:

We are asked about time and height at which both balls pass at the same height. So both, TIME and HEIGHT of both vertical trajectories are the same.

To find the values we use the kinematics expression for vertical motion with constant acceleration, using as the gravity acceleration -9.8 m/s2. Motion downwards is negative and upwards is positive. The reference point is the bottom of the building. The equation is as follow:

$y=y_0+v_0*t+0.5*a*t^2$

For each ball this equation is:

$y_A=28+4*t-4.9*t^2\\y_B=0+20*t-4.9*t^2$

Since the data of the problem is using SI units, then our answers will be expressed in SI units as well. Now, we first compute part B by equaling both equations (same height) and solving for time:

$y_A=y_B\\28+4*t-4.9*t^2=20*t-4.9*t^2\\-16t=-28\\t=1.750s$

With this time we can find the Height by substituting it in any equation of the balls. In this case, we use the expression of ball B

$y_B=20*1.75-4.9*1.75^2\\y_B=y_A=19.994m$