For a science fair competition, a group of high school students built a kicker -machine that can launch a golf ball from the origin with a velocity of 11.2m/s and initial angle of 31.5° with respect to the horizontal.
a) Where will the golf ball fall back to the ground?
b) How high will it be at the highest point of it's trajectory?
c) What is the ball's velocity vector (in Cartesian components) at the highest point of it's trajectory?
If you want to use a catapult to throw rocks and the
maximum range you need these projectiles to have is 0.67km , what initial speed do your projectiles have to have as they leave the catapult?
Look at this question answered by Damon over 5 years ago, just change the numbers to fit your question.
http://www.jiskha.com/display.cgi?id=1296412513
thanks
but why 1st two question answer? ??
please answer me.
sir!....
wating me....(°_^)
if you want to use a catapult to throw rocks and the maximum range you need these projectile to have is 0.67km what is the initial speed of the rocks
a) To determine where the golf ball will fall back to the ground, we need to find the range of the projectile. The range is the horizontal distance traveled by the golf ball before hitting the ground.
We can use the equations of motion to solve for the range. The horizontal range (R) can be calculated using the formula:
R = (v^2 * sin(2θ)) / g
Where:
- v is the initial velocity of the golf ball (11.2 m/s)
- θ is the launch angle (31.5°)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the values into the equation:
R = (11.2^2 * sin(2 * 31.5°)) / 9.8
Calculating the value gives us:
R ≈ 20.1 meters
Therefore, the golf ball will fall back to the ground approximately 20.1 meters away from the origin.
b) To determine the maximum height of the golf ball's trajectory, we can use the equation for the vertical displacement (y) of a projectile at its highest point:
y = (v^2 * sin^2(θ)) / (2 * g)
Substituting the given values:
y = (11.2^2 * sin^2(31.5°)) / (2 * 9.8)
Calculating the value gives us:
y ≈ 4.07 meters
Therefore, the golf ball will reach a maximum height of approximately 4.07 meters above the ground.
c) At the highest point of the trajectory, the vertical component of the ball's velocity is zero, but the horizontal component remains unchanged throughout the motion.
The horizontal component of the ball's velocity (Vx) can be found using the formula:
Vx = v * cos(θ)
Substituting the given values:
Vx = 11.2 * cos(31.5°)
Calculating the value gives us:
Vx ≈ 9.60 m/s
The vertical component of the ball's velocity (Vy) is zero at the highest point of the trajectory.
Therefore, the ball's velocity vector at the highest point of its trajectory is approximately (9.60 m/s, 0 m/s).
For the second question, to find the initial speed required for the projectiles to have a maximum range of 0.67 km, we can use the formula for range (R) mentioned earlier and solve for the initial velocity (v).
R = (v^2 * sin(2θ)) / g
Since the range is given in kilometers, we need to convert it to meters:
R = 0.67 km * 1000 = 670 meters
Rearranging the formula and solving for v:
v = sqrt((R * g) / sin(2θ))
Substituting the known values:
v = sqrt((670 * 9.8) / sin(2θ))
To find the initial speed, we also need to know the launch angle (θ) of the projectiles from the catapult. Please provide the launch angle, and I can calculate the initial speed required.