A tennis player strikes a ball when it is 1.1 m above the ground, with a velocity 20 m/s at 17 degrees above the horizontal. The net is 8 m away horizontally and it is 0.5 m high. How high, vertically, above the net does the ball pass?
i got 2.19.. is this correct?
That's what I've got too, good work!
0.97
To find out how high, vertically, above the net the ball passes, we can use the principles of projectile motion. Here's how:
1. Resolve the initial velocity into horizontal and vertical components:
- The horizontal component is given by: Vx = V * cos(θ), where V is the initial velocity (20 m/s) and θ is the angle (17 degrees).
- The vertical component is given by: Vy = V * sin(θ), where V is the initial velocity (20 m/s) and θ is the angle (17 degrees).
2. Calculate the time it takes for the ball to reach the net horizontally:
- We can use the formula: time = distance / velocity, where the distance is the horizontal distance to the net (8 m) and the velocity is the horizontal component of the initial velocity (Vx).
3. Determine the vertical displacement of the ball at the time it reaches the net:
- We can use the formula: displacement = Vy * time + (0.5 * g * t^2), where:
- Vy is the vertical component of the initial velocity,
- time is the time calculated in step 2,
- g is the acceleration due to gravity (approximately 9.8 m/s^2),
- t is the time.
4. Add the height of the net (0.5 m) to the vertical displacement calculated in step 3 to get the final answer.
Now, let's calculate the answer step by step:
Step 1:
Vx = 20 m/s * cos(17 degrees) ≈ 18.79 m/s
Vy = 20 m/s * sin(17 degrees) ≈ 5.31 m/s
Step 2:
time = 8 m / 18.79 m/s ≈ 0.425 seconds
Step 3:
displacement = 5.31 m/s * 0.425 seconds + (0.5 * 9.8 m/s^2 * (0.425 seconds)^2) ≈ 2.52 m
Step 4:
Final answer = 2.52 m + 0.5 m = 3.02 m
Therefore, the ball passes approximately 3.02 meters above the net vertically.