Question

From the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 7.60 m/s and angle of 18.0° below the horizontal. It strikes the ground 6.00 s later. What is the height y0?

Answer 1

To find the initial height, we can use the equation y = y0 + v0y * t + (1/2) * g * t^2, where y0 is the initial height, v0y is the vertical component of the initial velocity, t is the time of flight, and g is the acceleration due to gravity.

Step-by-step explanation:

To find the initial height, we need to find the displacement in the vertical direction. We can use the equation

y = y0 + v0y * t + (1/2) * g * t^2

where y0 is the initial height, v0y is the vertical component of the initial velocity, t is the time of flight, and g is the acceleration due to gravity. In this case, we know that y = 0 (since the ball hits the ground), v0y = v0 * sin(angle), and t = 6.00 s. Plugging in these values, we can solve for y0.