12. A ball is kicked with an initial velocity if 40 m/s from the ground at an angle of 30° to the
horizontal. (Use g= 10m/s²) Calculate
a) Horizontal and vertical component of initial velocity
b) The vertical velocity after t = 1s,
c) Position (r = xi + yj) aftert= 1s,
d) total time of flight (ime)
e) maximum height
f) Range of projectile
a) The horizontal component of the initial velocity can be found using the formula Vx = Vcosθ, where V is the magnitude of the initial velocity and θ is the angle of the initial velocity. So, Vx = 40 m/s * cos(30°) = 40 m/s * √3/2 ≈ 34.64 m/s.
The vertical component of the initial velocity can be found using the formula Vy = Vsinθ. So, Vy = 40 m/s * sin(30°) = 40 m/s * 1/2 = 20 m/s.
b) The vertical velocity after t = 1s can be found using the equation Vy = Vy0 - gt, where Vy0 is the initial vertical velocity and g is the acceleration due to gravity. So, Vy = 20 m/s - 10 m/s² * 1s = 20 m/s - 10 m/s = 10 m/s.
c) To find the position (r = xi + yj) after t = 1s, we can use the equations x = x0 + Vx0t and y = y0 + Vy0t - 1/2gt², where x and y are the position components at time t, x0 and y0 are the initial position components, Vx0 and Vy0 are the initial velocity components, and g is the acceleration due to gravity.
Since the ball is kicked from the ground, y0 = 0. Plugging in the known values, we get x = 0 + 34.64 m/s * 1s = 34.64 m and y = 0 + 20 m/s * 1s - 1/2 * 10 m/s² * (1s)² = 20 m - 5 m = 15 m.
Therefore, the position after t = 1s is r = 34.64i + 15j.
d) The total time of flight (ime) can be found using the equation t = (2 * Vy0) / g, where Vy0 is the initial vertical velocity and g is the acceleration due to gravity. Plugging in the known values, we get t = (2 * 20 m/s) / 10 m/s² = 40 m/s / 10 m/s² = 4s.
Therefore, the total time of flight is 4s.
e) The maximum height can be found using the equation H = (Vy0)² / (2 * g), where Vy0 is the initial vertical velocity and g is the acceleration due to gravity. Plugging in the known values, we get H = (20 m/s)² / (2 * 10 m/s²) = 400 m²/s² / 20 m/s² = 20 m.
Therefore, the maximum height is 20m.
f) The range of the projectile can be found using the equation R = Vx0 * t, where Vx0 is the initial horizontal velocity and t is the time of flight. Plugging in the known values, we get R = 34.64 m/s * 4s = 138.56 m.
Therefore, the range of the projectile is 138.56m.