Air Flight Unit

How long is the ball “in flight”? Answer in units of s.?

A ball is thrown from the top of a building upward at an angle of 43◦ to the horizontal and with an initial speed of 30 m/s, as in the figure. The ball is thrown at a height of 27 m above the ground. The acceleration of gravity is 9.8 m/s2 .How long is the ball “in flight”? Answer in units of s. What is the speed of the ball just before it strikes the ground? Answer in units of m/s.

Public Comments

1. Step 1: if you cant figure out where to start, draw a picture. jot down the information you are given and what you need to find out. in a chart on the side list your y and x values Y values 9.8m/s2 dy= 27 m step 2: so you got to find your Vy initial and Vx initial. seperate your x and y values X 30cos43 ---> V inital x----> 21.9m/s Y 30sin43 ---> V inital y---->20.5m/s Step 2: using the information you have found solve for time? the equation that can be used is Vf=V0 + at rewrite it solving for time---> Vf-V0/a = t your Vy initial is 20.5m/s your acceleration is -9.8m/s your Vf is 0 m/s because we are solving for half the parabola, in the end you will have to double the time you find through this equation. we are doing this because it is easier to find time using the final velocity of 0 m/s looking at the middle of your parabola (the highest point) at that point Vy inital is equivaltent to 0 m/s. 0m/s-20.5m/s all divided by - 9.8m/s2 =time note* negative accelartion being used because the ball is being thrown upwards whereas if it were to be thrown downwards it would be a positive acceleration. the time you get is 2.09 seconds multiply the time by two and you get 4.2 seconds. therefore the ball is in flight for 4.2 seconds. What is the speed of the ball just before it strikes the ground? Vfy= Vo +at Vfy= 0+(9.8)(4.2) Vfy= 41.2 m/s
2. a) Taking UP as the positive direction s = ut + 0.5 at^2 where: s = displacement (difference between the initial and final positions) = – 27 metres u = vertical component of initial velocity = 30 * sin 43 = 20.4600 m/s a = acceleration = – 9.8 m/s/s t = time taken in seconds Rearranging to give: 0.5at^2 + ut – s = 0 then the quadratic formula gives: t = – u / a ± √ (u^2 / a^2 + 2s / a) = – 20.46 / – 9.8 ± √ (20.46^2 / (– 9.8)^2 + 2 * – 27 / – 9.8) = 2.0878 ± 3.1415 = 5.2293 seconds (or – 1.0537) b) Now, considering velocity downwards i.e. DOWN as the positive direction h = (u * sin θ)^2 / 2a where: h = peak height of ball trajectory in metres u = initial velocity = 30 m/s θ = firing angle = 43 degrees a = acceleration = 9.8 m/s/s h = (30 * sin 43)^2 / (2 * 9.8) = 21.3576 m v^2 = u^2 + 2as where: v = final velocity in m/s u = initial velocity = 0 m/s as ball is stationary momentarily at peak height s = displacement from peak height to ground = 21.3576 + 27 = 48.3576 m v = √ (2 * a * s) = √ (2 * 9.8 * 48.3576) = 30.7865 m/s