To find the angle (θ) and the initial velocity (v) of the ball, we can use the principles of projectile motion.
Step 1: Separate the horizontal and vertical components of the ball's motion:
- The horizontal component only involves the ball's initial velocity since there are no horizontal accelerations: vx = v * cos(θ)
- The vertical component involves both the initial vertical velocity and the acceleration due to gravity: vy = v * sin(θ) - gt, where g is the acceleration due to gravity (-9.8 m/s^2).
Step 2: Calculate the time it takes for the ball to reach the ground (t):
- In this case, the given time is t = 3.2 s.
Step 3: Use the vertical motion equation to determine the initial vertical velocity (vy):
- We can use the equation: 0 = vy - gt
- Since the ball lands on the ground, the final height is 0, and we can substitute the values: 0 = v * sin(θ) - 9.8 * 3.2
Step 4: Calculate the initial horizontal velocity (vx):
- Since the horizontal component of motion is not affected by gravity, the horizontal velocity (vx) remains constant.
- Since the ball lands 10m away from the building, the horizontal displacement (dx) is 10m.
- We can use the equation: dx = vx * t
- Substituting the values: 10 = v * cos(θ) * 3.2
Step 5: Solve for vy and vx using the equations from Step 3 and Step 4:
- From the equation in Step 3: v * sin(θ) = 9.8 * 3.2
- From the equation in Step 4: v * cos(θ) = 10 / 3.2
Step 6: Solve for v and θ by dividing the two equations obtained in Step 5:
- Dividing the equation v * sin(θ) = 9.8 * 3.2 by the equation v * cos(θ) = 10 / 3.2 results in:
- tan(θ) = (9.8 * 3.2) / (10 / 3.2)
- tan(θ) = 9.8 * 3.2 * (3.2 / 10)
- tan(θ) = 9.8 * (3.2)^2 / 10
Step 7: Calculate θ by taking the inverse tangent (arctan) of both sides:
- θ = arctan(9.8 * (3.2)^2 / 10)
Step 8: Calculate v by dividing the equation from Step 4 by cos(θ):
- v = 10 / (3.2 * cos(θ))
Substituting the value of θ into the equation in Step 8 will give you the final value of v.