Can someone just lead me in the right direction please

[Bucky8624]

1. A football is kicked with an initial speed of 22m/s at some angle above the horizontal. It travels 45 meters horizontally before hitting the ground.
a. At what angle was the ball initially kicked?
b. How long does it stay in the air?

 

[Hypercase]

Split the 22m/s into vertical ( Vy = 22*sin theta) and horizontal components
(Vx =22*Cos theta). Now use the fact that Vy at the highest pt is 0.(use 0 = 22 sin theta - 9.8*t1).
also since motion in the horizontal direction is uniform motion
45 = 22 Cos theta / t.
Also the time taken to reach the highest pt. is half the time taken to reach the ground. ==> t = 2*t1.

 

[quicknote]

Sure, I can help guide you in the right direction. To solve this problem, we can use the kinematic equations for projectile motion. The first step would be to break down the initial velocity of 22m/s into its horizontal and vertical components. We can use trigonometry to find that the initial vertical velocity is 22sinθ, where θ is the angle at which the ball was kicked.

a. To find the angle, we can use the equation tanθ = (22sinθ)/(22cosθ), which simplifies to tanθ = sinθ/cosθ. Using a calculator, we can find that the angle is approximately 23.6 degrees.

b. To find the time the ball stays in the air, we can use the equation d = vt + (1/2)at^2, where d is the horizontal distance traveled (45m), v is the horizontal component of the initial velocity (22cosθ), a is the acceleration due to gravity (-9.8m/s^2), and t is the time. Rearranging the equation, we get t = (-v ± √(v^2 - 4(1/2)a(-d)))/2(1/2)a. Substituting in the values, we get t = (-22cosθ ± √((22cosθ)^2 - 4(1/2)(-9.8)(-45)))/2(1/2)(-9.8). Using the angle we found in part a, we can solve for t and find that the ball stays in the air for approximately 3.3 seconds.

I hope this helps guide you in the right direction. Let me know if you have any further questions.