Question

On a distant planet the equation y= -5t2+v0t+h0represents the trajectory of a thrown object where v0represents the upward initial velocity and h0represents the initial height. A tall basketball player throws the ball at 40 feet high, with an upward initial velocity of 10 feet per second. How long is the ball in the air? Solve by compleating the square.

Answer 1

To determine the time a basketball is in the air when thrown, we use the equation for projectile motion on a distant planet and complete the square. Given the initial conditions, we find that the ball is in the air for 4 seconds.

Step-by-step explanation:

The student is asked to calculate how long a basketball is in the air when thrown on a distant planet. The equation provided is y = -5t2 + v0t + h0, where v0 is the initial velocity and h0 represents the initial height.

Given that the initial velocity (v0) is 10 feet per second and the initial height (h0) is 40 feet, we plug into the equation to get y = -5t2 + 10t + 40. To solve for the time (t) when the ball hits the ground (y = 0), we complete the square:

1. Move h0 to the other side: 0 = -5t2 + 10t + 40.
2. Divide all terms by -5 (to make the coefficient of t2 equal to 1): 0 = t2 - 2t - 8.
3. Rearrange the equation: t2 - 2t = 8.
4. Add (-2/2)2 = 1 to both sides to complete the square: t2 - 2t + 1 = 9.
5. Factor the left side: (t - 1)2 = 9.
6. Take the square root of both sides:

t - 1 = ± 3

• t = 4 seconds (throwing upwards)
• t = -2 seconds (not physically meaningful in this context)

Hence, the ball is in the air for 4 seconds before it hits the ground.