Suppose a ball is thrown directly upward from a height of 7 feet with an initial velocity of 50 feet per second. Use the quadratic formula or a graphing calculator to find the number of seconds it takes the ball to hit the ground. Round the nearest tenth of a second.

Accepted Solution

Answer:t = 3.3 secondsStep-by-step explanation:From the formula of vertical motion of an object under gravity we can write the equation[tex]H = ut + \frac{1}{2} gt^{2}[/tex] ....... (1)Where u is the initial velocity (in feet per second) of throw of the object and t is time of travel in seconds and the value of g i.e. gravitational acceleration is 32 feet/sec².Now, while a ball is thrown vertically upward with velocity 50 ft/sec from a height of 7 ft then the time of travel of the ball before reaching the ground, the equation (1) will be written as[tex]- 7 = 50t - \frac{1}{2} \times 32 \times t^{2}[/tex]As we have selected the upward direction as positive so, gravitational acceleration,g will be negative and as the displacement is downward by 7 feet, so it will be negative.⇒ 16t² - 50t - 7 = 0 ........ (2)Now, applying Sridhar Acharya formula,[tex]t = \frac{-(-50) + \sqrt{(-50)^{2} - 4(16)(-7)}}{2(16)}[/tex] {Neglecting the negative root as t can not be negative}⇒ t = 3.3 seconds {Rounded to the nearest tenth}(Answer)