A ball is thrown so that its motion is defined by the equations:
x = 5t; y = 2 + 6t - 4.9t2, where and x and y are in meters and t is in seconds. Determine:
1. The velocity of the ball at any time t
2. The initial position and velocity of the ball
3. The velocity of the ball at time = 1.1 second
4. The horizontal distance the ball travels before it hits the ground.
5. The time at which the ball hits the ground
6. The magnitude and direction of the velocity vector when the ball hits the ground
#1. v = 5i + (6-9.8t)j
#2. set t=0
#5 find t when y=0
the rest should be no trouble
To determine the answers, we will work step by step:
1. The velocity of the ball at any time t:
The velocity of the ball is given by the derivatives of x and y with respect to time t. Taking the derivative of x = 5t, we get:
vx = d/dt(5t) = 5 m/s
Similarly, taking the derivative of y = 2 + 6t - 4.9t^2, we get:
vy = d/dt(2 + 6t - 4.9t^2) = 6 - 9.8t m/s
Therefore, the velocity of the ball at any time t is given by the vector (5, 6 - 9.8t) m/s.
2. The initial position and velocity of the ball:
The initial position of the ball corresponds to t = 0. Substituting t = 0 into the equations, we find:
x(0) = 5(0) = 0 m
y(0) = 2 + 6(0) - 4.9(0)^2 = 2 m
Therefore, the initial position of the ball is (0, 2) m.
The initial velocity of the ball can be found by substituting t = 0 into the velocity equations:
vx(0) = 5 m/s
vy(0) = 6 - 9.8(0) = 6 m/s
Therefore, the initial velocity of the ball is (5, 6) m/s.
3. The velocity of the ball at time t = 1.1 second:
Substituting t = 1.1 into the velocity equations, we get:
vx(1.1) = 5 m/s
vy(1.1) = 6 - 9.8(1.1) = -4.58 m/s
Therefore, the velocity of the ball at time t = 1.1 seconds is (5, -4.58) m/s.
4. The horizontal distance the ball travels before it hits the ground:
To find the horizontal distance, we need to know the time at which the ball hits the ground. The ball hits the ground when y = 0. So, we can set y = 0 in the equation and solve for t:
0 = 2 + 6t - 4.9t^2
Solving for t using the quadratic formula, we find two solutions, t = 0.41 s and t = 2.1508 s. Since we are interested in the time before the ball hits the ground, we only consider the positive solution, t = 2.1508 s.
Therefore, the horizontal distance the ball travels before it hits the ground is given by substituting t = 2.1508 into the x equation:
x(2.1508) = 5(2.1508) = 10.754 m
So, the ball travels approximately 10.754 meters horizontally before hitting the ground.
5. The time at which the ball hits the ground:
As mentioned earlier, the ball hits the ground when y = 0. From the quadratic equation used in the previous step, we found the time at which y = 0 as t = 2.1508 s.
Therefore, the ball hits the ground at approximately t = 2.1508 seconds.
6. The magnitude and direction of the velocity vector when the ball hits the ground:
To find the magnitude of the velocity vector when the ball hits the ground, we can calculate the velocity at t = 2.1508 seconds using the velocity equations:
vx(2.1508) = 5 m/s
vy(2.1508) = 6 - 9.8(2.1508) = -10 m/s
The magnitude of the velocity vector can be calculated using the Pythagorean theorem:
|v| = sqrt(vx^2 + vy^2)
|v| = sqrt((5)^2 + (-10)^2) = sqrt(125) = 11.1803 m/s
Therefore, the magnitude of the velocity vector when the ball hits the ground is approximately 11.1803 m/s.
To determine the direction of the velocity vector, we need to calculate the angle it makes with the horizontal. The angle is given by:
θ = tan^(-1)(vy/vx)
θ = tan^(-1)(-10/5) = -63.43° (approximately)
Therefore, the direction of the velocity vector when the ball hits the ground is approximately -63.43° with respect to the horizontal.
To answer these questions, we need to understand the equations and formulas related to motion. Let's go step by step:
1. The velocity of the ball at any time t:
Velocity (V) is the rate of change of position with respect to time. It can be determined by taking the derivative of the position equation with respect to time.
Given: x = 5t; y = 2 + 6t - 4.9t^2
To find the velocity, differentiate the position equations with respect to time:
Vx = dx/dt = d(5t)/dt = 5 (since the derivative of t is 1)
Vy = dy/dt = d(2 + 6t - 4.9t^2)/dt = 6 - 9.8t
So, the velocity of the ball at any time t is V = (Vx, Vy) = (5, 6 - 9.8t)
2. The initial position and velocity of the ball:
At t = 0, we can find the initial position and velocity.
For position:
x = 5t = 5(0) = 0 m
y = 2 + 6t - 4.9t^2 = 2 + 6(0) - 4.9(0)^2 = 2 m
So, the initial position is (0, 2) meters.
For velocity:
Using the velocity equation obtained earlier:
V = (Vx, Vy) = (5, 6 - 9.8(0)) = (5, 6)
So, the initial velocity is (5, 6) m/s.
3. The velocity of the ball at time = 1.1 second:
Substitute t = 1.1 into the velocity equation.
V = (5, 6 - 9.8(1.1)) = (5, 6 - 10.78) = (5, -4.78)
So, the velocity of the ball at t = 1.1 seconds is (5, -4.78) m/s.
4. The horizontal distance the ball travels before it hits the ground:
To find the horizontal distance, we need to determine the time when the ball hits the ground.
At the ground, y = 0.
0 = 2 + 6t - 4.9t^2
To solve for t, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = -4.9, b = 6, c = 2.
t = [ -6 ± √(6^2 - 4(-4.9)(2)) ] / (2(-4.9))
t ≈ [ -6 ±√(36 + 39.2) ] / (-9.8)
t ≈ [ -6 ±√75.2 ] / (-9.8)
Since time cannot be negative for this scenario, we only consider the positive solution:
t ≈ ( -6 +√75.2 ) / (-9.8)
Substituting this value of t into the x equation to find the horizontal distance:
x = 5t ≈ 5[( -6 +√75.2 ) / (-9.8)]
Therefore, the horizontal distance the ball travels before hitting the ground is approximately 2.13 meters.
5. The time at which the ball hits the ground:
As calculated earlier, the equation becomes:
t ≈ ( -6 +√75.2 ) / (-9.8)
Evaluating this expression gives approximately t ≈ 1.05 seconds.
So, the ball hits the ground at approximately 1.05 seconds.
6. The magnitude and direction of the velocity vector when the ball hits the ground:
At the moment the ball hits the ground, the velocity vector will have both horizontal and vertical components.
Substituting t = 1.05 into the velocity equation gives:
V = (5, 6 - 9.8(1.05)) = (5, -3.69)
The magnitude of the velocity can be determined using the Pythagorean theorem:
|V| = √(Vx^2 + Vy^2) ≈ √(5^2 + (-3.69)^2) ≈ √(25 + 13.5961) ≈ √38.5961 ≈ 6.21 m/s
The direction of the velocity vector can be determined using trigonometry:
θ = tan^(-1)(Vy / Vx) ≈ tan^(-1)(-3.69 / 5) ≈ -38.66°
Therefore, the magnitude of the velocity vector is approximately 6.21 m/s, and the direction is approximately -38.66° (measured counterclockwise from the positive x-axis).