A ball is thrown from ground level at an angle of 0 and the initial velocity components are 7.5m/s and 13m/s. Ignoring air resistance, calculate each of: 1- The initial velocity is vi

2- The value of the angle at which the ball was thrown

3 - The horizontal range of the ball R

I assume you mean an angle of θ, and not 0. I also assume the given components of the initial velocity are horizontal and vertical, respectively, so that

1. the initial velocity vector is

which means the ball is thrown with an initial speed of

√((7.5 m/s)² + (13 m/s)²) ≈ 15 m/s,

and

2. the angle made by with the positive horizontal axis is θ such that

3. At time t, the ball attains a height y and horizontal range x according to

y = (13 m/s) t - g/2 t²

x = (7.5 m/s) t

where g = 9.8 m/s². When the ball reaches the ground (y = 0) for t > 0, we have

(13 m/s) t - g/2 t² = 0

13 m/s - g/2 t = 0

t = 2 (13 m/s)/g = (26 m/s)/g

Plugging this time into the x equation gives a horizontal range of

x = (7.5 m/s) (26 m/s)/g ≈ 20. m

power out to power in less than 100%

neither accurate nor precise.