Andrew K. answered • 08/17/20
Student-Athlete and Physics/Computer Science Double Major at MIT
Using the starting height of 11 meters and starting velocity of 7 m/s, we can write the equation for the height of the ball at time t as h(t) = 11 + 7t - 4.9t^2. It is important to note that this equation only applies for t greater than or equal to zero because we are starting at height 11 meters at time equal to zero and time cannot go backwards.
a. Using the discriminant (b^2 - 4ac = 49 + 4 * 4.9 * 11 = 264.6), because the discriminant is positive, we see that there are two solutions.
b. If we graph the height equation, we can verify that the equation has two solutions where the height is zero. However, it is important to note that one of these solutions has a negative value, so that solution does not apply to this physical scenario because time cannot be negative.
c. Using factoring, graphing, or the quadratic formula, we will find that the height is zero at two different times: t = -0.946 seconds and t = 2.37 seconds.
d. Each method gives the same answer which should always be true. However, factoring is significantly more difficult than the quadratic formula for this problem.
e. The solutions mean that the ball is at height zero at 2.37 seconds after the ball is thrown and 0.946 seconds before the ball is thrown. However, because time cannot be negative, the only solutions that apply are one which have t greater than or equal to zero. So the only solution that applies is t = 2.37 seconds, so the ball will hit the ground after 2.37 seconds.
f. The domain of the problem is t greater than or equal to zero, because the problem starts at t equal to zero and time always increases, so negative time does not make sense in this problem.
