A ball launched from ground level lands 2.15 s later on a level field 50 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.)
u = constant horizontal speed = 50/2.5 = 20 m/s
v = vi - 9.8 t
at top of arc, t = 2.5/2 = 1.25 seconds and v = 0
so at top
0 = vi - 9.8 (1.25)
so
vi = 12.25
so speed at start = sqrt (20^2 + 12.25^2)
= sqrt(400+ 150) = sqrt 550 = 23.5 m/s
tan angle = 12.25/20
so
angle = 31.5 degrees
To find the magnitude of the initial velocity vector and the angle it is above the horizontal, we can use the equations of motion in two dimensions. Here's how you can solve it step by step:
Step 1: Break down the initial velocity vector into its horizontal and vertical components. Denote the magnitude of the initial velocity as v and the angle it makes with the horizontal as θ. The horizontal component of the initial velocity (v_x) can be calculated as v_x = v * cos(θ), and the vertical component (v_y) can be calculated as v_y = v * sin(θ).
Step 2: Since the ball lands 2.15 seconds later and moves 50 meters horizontally, we can use the equation for horizontal displacement: x = v_x * t, where x is the horizontal displacement and t is the time. In this case, x = 50 meters and t = 2.15 seconds.
Step 3: Rearrange the equation x = v_x * t to solve for v_x: v_x = x / t. Substituting the given values, we have v_x = 50 m / 2.15 s.
Step 4: Now, we need to find the vertical component of the velocity (v_y) at the time of landing. Since the ball is in free fall vertically, we can use the equation for vertical displacement: y = v_y * t + (1/2) * g * t^2, where y is the vertical displacement and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Step 5: Since the ball is launched from ground level, the initial vertical displacement (y) is 0. Therefore, the equation simplifies to 0 = v_y * t + (1/2) * g * t^2. We substitute v_y = v * sin(θ), t = 2.15 s, and g = 9.8 m/s^2.
Step 6: Solve the equation 0 = v * sin(θ) * t + (1/2) * g * t^2 for v. Rearrange the equation to solve for v: v = -[(1/2) * g * t^2] / (sin(θ) * t).
Step 7: Substitute the given values of g = 9.8 m/s^2, t = 2.15 s, and θ into the equation to determine the magnitude of the initial velocity (v). Calculate v using the equation obtained in Step 6.
Step 8: To find the angle θ, use the equation tan(θ) = v_y / v_x. Substitute the values of v_y and v_x, calculated in Step 1 and Step 3, respectively, and solve for θ.
By following these steps and performing the necessary calculations, you can find the magnitude of the initial velocity vector and the angle it is above the horizontal.