. A ball is thrown so that the motion is
defined by the equations x = 5t and y = 2 +
6t – 4.9t2
, where x and y are expressed in
meters and t is expressed in seconds.
Determine (a) the velocity at t =1 s, (b) the
horizontal distance the ball travels before
hitting the ground.
(a)dy/dt = 6-9.8t; dx/dt = 5
Now use those to get the components of v at t=1
(b) find t when y(t)=0, then use that in x(t)
To determine the velocity at t = 1s, we need to find the derivatives of the x and y equations with respect to time and evaluate them at t = 1s.
Given:
x = 5t
y = 2 + 6t - 4.9t^2
Step 1: Find the derivative of x with respect to t.
Use the power rule: d/dt (c*t^n) = n*c*t^(n-1)
d/dt (5t) = 5*1*t^(1-1) = 5
Step 2: Find the derivative of y with respect to t.
d/dt (2 + 6t - 4.9t^2) = 6 - 2*4.9t = 6 - 9.8t
Step 3: Evaluate the derivatives at t = 1s.
The velocity at t = 1s is the derivative of x, which equals 5, and the derivative of y, which equals 6 - 9.8*1 = -3.8.
(a) The velocity at t = 1s is 5 m/s horizontally and -3.8 m/s vertically.
To find the horizontal distance traveled by the ball before hitting the ground, we need to find the time when y = 0 and substitute it back into the x equation.
Step 1: Set y = 0 and solve for t.
2 + 6t - 4.9t^2 = 0
-4.9t^2 + 6t + 2 = 0
Step 2: Solve the quadratic equation for t.
Using the quadratic formula: t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = -4.9, b = 6, and c = 2.
t = (-6 ± sqrt(6^2 - 4(-4.9)(2))) / (2*-4.9)
t = (-6 ± sqrt(36 + 39.2)) / (-9.8)
t = (-6 ± sqrt(75.2)) / (-9.8)
Using a calculator, we find two solutions: t ≈ 0.39s and t ≈ 1.29s.
Step 3: Substitute t = 1.29s into the x equation.
x = 5t
x = 5 * 1.29
x ≈ 6.45
(b) The horizontal distance traveled by the ball before hitting the ground is approximately 6.45 meters.