A ball is thrown directly upward with an initial velocity of 15 m/s from the rooftop of a building that is 30 meters in height. It is caught by a student whose hand is extended from a window. If the students hand is located 18 meters below the rooftop. What is the elapsed time just before the ball reaches the students outstretched hand in seconds
To solve this problem, we can use the kinematic equation for vertical motion:
h = h0 + v0*t - 0.5*g*t^2
where:
h is the final height (18 meters below the rooftop)
h0 is the initial height (30 meters)
v0 is the initial velocity (15 m/s)
g is the acceleration due to gravity (9.8 m/s^2)
t is the elapsed time
Rearranging the equation, we get:
0.5*g*t^2 - v0*t + (h0 - h) = 0
Substituting the given values, the equation becomes:
0.5*9.8*t^2 - 15*t + (30 - 18) = 0
Simplifying the equation:
4.9*t^2 - 15*t + 12 = 0
Now we can solve this quadratic equation for t. Using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 4.9, b = -15, and c = 12. Substituting these values into the quadratic formula:
t = (15 ± sqrt((-15)^2 - 4*4.9*12)) / (2*4.9)
Simplifying the equation:
t = (15 ± sqrt(225 - 235.2)) / 9.8
t = (15 ± sqrt(-10.2)) / 9.8
Since the discriminant (-10.2) is negative, the quadratic equation has no real solutions. This means that the ball will not reach the student's hand.
Therefore, the elapsed time just before the ball reaches the student's outstretched hand is undefined.