A ball is thrown so that the motion is defined by the equations x=5t and y=2+6t-4.9t^2, where x and y are expressed in meters and t is expressed in seconds. a) find velocity at t=1s, b) find the horizontal distance the ball travels before hitting the ground.
solutions
a) To find the velocity at t=1s, we need to find the derivative of the position equation with respect to time.
The given equation for x is x = 5t, which means the velocity in the x direction is the derivative of x with respect to t.
Taking the derivative, we get:
x' = d/dt (5t) = 5
So, the velocity in the x direction is constant and equal to 5 m/s.
Now, let's find the velocity in the y direction. The given equation for y is y = 2 + 6t - 4.9t^2.
Taking the derivative, we get:
y' = d/dt (2 + 6t - 4.9t^2) = 6 - 9.8t
So, the velocity in the y direction is given by vy = 6 - 9.8t.
Substituting t=1s into the equation, we get:
vy = 6 - 9.8(1)
vy = 6 - 9.8
vy = -3.8 m/s
Therefore, the velocity at t=1s is 5 m/s in the x direction and -3.8 m/s in the y direction.
b) To find the horizontal distance the ball travels before hitting the ground, we need to determine the time when the ball hits the ground. The ball hits the ground when y=0.
Setting y=0 in the given equation:
0 = 2 + 6t - 4.9t^2
This is a quadratic equation. Let's solve it by setting it equal to zero and using the quadratic formula.
-4.9t^2 + 6t + 2 = 0
Using the quadratic formula, we get:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -4.9, b = 6, and c = 2.
Substituting the values into the quadratic formula, we have:
t = (-6 ± √(6^2 - 4(-4.9)(2))) / (2(-4.9))
Simplifying the expression, we get:
t = (-6 ± √(36 + 39.2)) / (-9.8)
t = (-6 ± √75.2) / (-9.8)
Using a calculator, we find two possible values for t: t ≈ 1.642s and t ≈ -0.241s.
Since time cannot be negative in this context, we discard t ≈ -0.241s.
Therefore, the ball hits the ground at t ≈ 1.642s.
To find the horizontal distance the ball travels, we can substitute this value of t into the equation for x: x = 5t.
x = 5(1.642)
x ≈ 8.21 meters.
Therefore, the ball travels approximately 8.21 meters horizontally before hitting the ground.
To answer these questions, we need to find the velocity at t=1s and the horizontal distance the ball travels before hitting the ground using the given equations.
a) Finding the velocity at t=1s:
Velocity can be calculated by taking the derivative of the position equation with respect to time. Applying this to the given equation x=5t, we get:
dx/dt = d(5t)/dt = 5.
So, the velocity at any given time t for the x-coordinate is 5 m/s.
Similarly, for the equation y=2+6t-4.9t^2, we need to find dy/dt. Differentiating y with respect to t, we have:
dy/dt = d(2+6t-4.9t^2)/dt = 6 - 2 * 4.9t.
Now, when t = 1s, dy/dt is:
dy/dt = 6 - 2 * 4.9 * 1 = 6 - 9.8 = -3.8.
Therefore, the velocity at t=1s for the y-coordinate is -3.8 m/s.
b) Finding the horizontal distance before hitting the ground:
The ball hits the ground when y=0 since the ground is at y=0. We can solve the equation y=2+6t-4.9t^2 for t by setting y=0:
0 = 2+6t-4.9t^2
This is a quadratic equation, which can be solved by factoring or using the quadratic formula. In this case, we will use the quadratic formula to find the roots:
t = (-b ± √(b^2 - 4ac)) / (2a),
where a = -4.9, b = 6, and c = 2.
Plugging these values into the formula, we get:
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 we are interested in the positive value of t, we can discard the negative solution. Evaluating the positive value using a calculator, we find:
t ≈ 1.62s.
Now, to find the horizontal distance traveled, substitute t = 1.62 into the x-coordinate equation:
x = 5t
x = 5(1.62)
x ≈ 8.1m.
Therefore, the horizontal distance the ball travels before hitting the ground is approximately 8.1 meters.