Physics Homework

Question

A ball launched from ground level lands 2.8 s later on a level field 43 m away from the launch point. Find the magnitude of the initial velocity vector and the angle it is above the horizontal. (Ignore any effects due to air resistance.)

Magnitude = ? m/s

direction = ? (in degrees)

Answer 1

Hi, Conor, let's see what we can do with this one. First draw a diagram representing the problem. Draw a horizontal line representing the distance the ball goes. Draw a small arrow at the left end of the line in a direction somewhat in between the x and y axes. This represents the initial velocity vector. Finally, draw a parabolic shaped line from the end of the vector to a point on the horizonal line which represents where the ball lands.

The initial velocity vector has two components, one in the x axis direction (Vx) and the other in the y axis direction (Vy). Vx is found by the following equation Vx = 43m / 2.8 s = 15.36 m/s. Vy is not so easy.

First we have to recognize that the time for the ball to reach its highest point is equal to 1/2 the total time or 1.4 seconds. One equation involving the vertical velocity is Vy/2 x time = max height or Vy/2 x 1.4 = h where Vy/2 is the average upward velocity as the final velocity going up is zero, and h is the maximum height. We still have too many unknowns, so we need another equation. A second equation we can use is

h = Vy x t - 1/2 x g x t². Now we can equate the two h equations, Vy/2 x 1.4 = Vy x t -1/2 x g x t². Now there is only one unknown, Vy because t = 1.4 sec and g = 9.8 m/s². If we substitute the known quantities, we get, with some algebraic manipulation, Vy = 13.71 m/s. Now we have enough information to calculate the initial velocity vector. The magnitude is obtained by the Pythagorean theorem Vinit. = √Vx² + Vy² = 20.6 m/s. The angle from the x axis is gotten by arctan (Vy / Vx) = arctan (13.71/ 15.36) = 41.8 degrees.

Please check my math as there were many different calculations.