A small ball is launched at an angle of 30.0 degrees about the horizontal. It reaches a maximum height of 2.5m with a respect to the launch position.
What is the initial velocity of the ball when its launched?
What is the range, defined as a horizontal distance traveled until it returns to his original height. (you can ignore air resistance)
To determine the initial velocity of the ball when it is launched, we can use the equation of motion for projectile motion. The equation for the maximum height reached by a projectile is given by:
H = (v^2 * sin^2θ) / (2g)
Where:
H is the maximum height reached
v is the initial velocity of the ball
θ is the launch angle
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Plugging in the given values:
H = 2.5 m
θ = 30 degrees
g = 9.8 m/s^2
We can rearrange the equation to solve for the initial velocity (v):
v = √((2g * H) / sin^2θ)
Substituting the values:
v = √((2 * 9.8 * 2.5) / sin^2(30))
Calculating,
v = √((49) / (0.25))
v = √(196)
v = 14 m/s
Therefore, the initial velocity of the ball when it is launched is 14 m/s.
To find the range of the ball, we can use the equation:
R = (v^2 * sin(2θ)) / g
Where:
R is the horizontal distance traveled until it returns to its original height
v is the initial velocity of the ball
θ is the launch angle
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Using the given values:
v = 14 m/s
θ = 30 degrees
g = 9.8 m/s^2
Substituting the values:
R = ((14^2) * sin(2 * 30)) / 9.8
Calculating,
R = (196 * 0.5) / 9.8
R = 98 / 9.8
R = 10 m
Therefore, the range of the ball, defined as the horizontal distance traveled until it returns to its original height, is 10 meters.
To find the initial velocity of the ball when it is launched, we can use the information given about the maximum height reached and the launch angle. Here's how you can calculate it:
Step 1: Identify the given information:
- Launch angle (θ) = 30.0 degrees
- Maximum height (h) = 2.5 m
- Acceleration due to gravity (g) = 9.8 m/s^2 (assuming Earth's gravity)
Step 2: Use the kinematic equation for maximum height:
The equation to calculate the maximum height reached by an object launched at an angle is:
h = (v^2 * sin^2(θ)) / (2 * g)
Rearrange the equation to solve for initial velocity (v):
v = √[(h * 2 * g) / sin^2(θ)]
Step 3: Plug in the values and calculate the initial velocity:
Substitute the given values into the equation.
v = √[(2.5 * 2 * 9.8) / sin^2(30.0)]
Calculating this expression will give you the initial velocity of the ball when it is launched.
To determine the range of the ball, defined as the horizontal distance traveled until it returns to its original height, we can use the equation for range. Here's how you can calculate it:
Step 1: Identify the given information:
- Launch angle (θ) = 30.0 degrees
- Initial velocity (v) obtained from the previous calculation
- Acceleration due to gravity (g) = 9.8 m/s^2 (assuming Earth's gravity)
Step 2: Use the kinematic equation for range:
The equation to calculate the range of a projectile launched at an angle is:
R = (v^2 * sin(2θ)) / g
Step 3: Plug in the values and calculate the range:
Substitute the given values into the equation.
R = [(v^2 * sin(60.0 degrees)) / 9.8]
Calculating this expression will give you the range of the ball.
Yf^2 = Yo^2 + 2g*h,
Yo^2 = Yf^2 - 2g*h,
Yo^2 = o - (-19.6)*2.5 = 960.4,
Yo = 31 m/s = Ver. component of Vo.
a. Vo = Yo/sinA = 31/sin30 = 62 m/s = Initial velocity.
b. R = Vo^2*sin(2A)/g,
R = (62)^2*sin60/9.8 = 339.7 m.