In a water balloon toss a contestant throws her balloon with a speed of 4.84 m/s at an angle of 16.5 degrees above the horizontal. Find the maximum height, the time of flight and the range of the balloon assuming her partner catches the balloon at the same height at which it was released.
time of flight:
hf=hi+Vi*sinTheta*t-g t^2
hf=hi find t.
now time to max height is half that time.
To solve this problem, we can use the kinematic equations of motion to analyze the motion of the water balloon.
First, we need to break down the initial velocity of the balloon into its horizontal and vertical components:
Vx = V * cos(θ)
Vy = V * sin(θ)
Given:
V = 4.84 m/s (speed of the balloon)
θ = 16.5 degrees (angle above the horizontal)
Calculating the components:
Vx = 4.84 m/s * cos(16.5°)
Vx = 4.84 m/s * 0.96126
Vx ≈ 4.65 m/s
Vy = 4.84 m/s * sin(16.5°)
Vy = 4.84 m/s * 0.27564
Vy ≈ 1.34 m/s
Now, let's solve for the maximum height:
The time taken to reach the maximum height is given by:
t_max_height = Vy / g
where g is approximately equal to 9.8 m/s^2 (acceleration due to gravity)
t_max_height = 1.34 m/s / 9.8 m/s^2
t_max_height ≈ 0.1367 seconds
The maximum height above the ground can be found using the equation:
h_max_height = (Vy^2) / (2 * g)
h_max_height = (1.34 m/s)^2 / (2 * 9.8 m/s^2)
h_max_height ≈ 0.0923 meters
Next, let's find the time of flight:
The total time of flight can be calculated using:
t_flight = 2 * t_max_height
t_flight = 2 * 0.1367 seconds
t_flight ≈ 0.2734 seconds
Finally, let's calculate the range of the balloon:
The range (horizontal distance traveled) can be determined by:
range = Vx * t_flight
range = 4.65 m/s * 0.2734 seconds
range ≈ 1.27 meters
Therefore, the maximum height of the balloon is approximately 0.0923 meters, the time of flight is approximately 0.2734 seconds, and the range of the balloon is approximately 1.27 meters.