In a friendly game of handball, you hit the ball essentially at ground level and send it toward the wall with a speed of 14.8 m/s at an angle of 34.1° above the horizontal. How long does it take for the ball to reach the wall if it is 4.10 m away? How high is the ball when it hits the wall?
To find the time it takes for the ball to reach the wall, we need to first determine the horizontal and vertical components of the ball's initial velocity.
The horizontal component of the velocity can be found using the equation:
Vx = V * cosθ
where Vx is the horizontal component of the velocity, V is the magnitude of the velocity (14.8 m/s in this case), and θ is the angle above the horizontal (34.1° in this case).
Plugging in the values:
Vx = 14.8 m/s * cos(34.1°)
Using a calculator, we find:
Vx ≈ 12.22 m/s
The vertical component of the velocity can be found using the equation:
Vy = V * sinθ
where Vy is the vertical component of the velocity, V is the magnitude of the velocity (14.8 m/s in this case), and θ is the angle above the horizontal (34.1° in this case).
Plugging in the values:
Vy = 14.8 m/s * sin(34.1°)
Using a calculator, we find:
Vy ≈ 8.08 m/s
Now, we can use the equation for vertical motion to find the time it takes for the ball to reach the wall:
y = Vy * t + (1/2) * g * t^2
where y is the vertical displacement of the ball, Vy is the initial vertical component of the velocity, t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the ball starts and ends at ground level, the vertical displacement is equal to zero. Therefore, we can rewrite the equation as:
0 = Vy * t - (1/2) * g * t^2
Rearranging the equation, we have:
(1/2) * g * t^2 = Vy * t
Dividing both sides by t and rearranging, we get:
(1/2) * g * t = Vy
Substituting the known values, we have:
(1/2) * 9.8 m/s^2 * t ≈ 8.08 m/s
Simplifying, we find:
4.9 m/s^2 * t ≈ 8.08 m/s
Dividing both sides by 4.9 m/s^2, we find:
t ≈ 1.65 s
Therefore, it takes approximately 1.65 seconds for the ball to reach the wall.
To find the height of the ball when it hits the wall, we can use the equation for vertical motion:
y = Vy * t + (1/2) * g * t^2
Substituting the known values, we have:
y = 8.08 m/s * 1.65 s + (1/2) * 9.8 m/s^2 * (1.65 s)^2
Calculating the values, we find:
y ≈ 13.36 m
Therefore, the ball is approximately 13.36 meters high when it hits the wall.