On planet Calculi, acceleration due to gravity is -20 ft/sec^2. A ball is thrown upward with an initial velocity of 40ft/sec from an initial height of 50.
a) Write the equation for the height of the ball, s(t), at time t.
b) Find the velocity when the ball hits the ground.
c) Find the time when the ball hits its maximum height.
I know velocity is the derivative and acceleration is the second derivative, but I'm getting confused on notation and solving. Thanks.
h(t) = 50 + 40t - 10t^2
The rest is easy.
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Right?
To answer these questions, we will use the following equations of motion in one dimension:
1. Velocity equation: v(t) = v0 + a*t
2. Position equation: s(t) = s0 + v0*t + 0.5*a*t^2
Here, v(t) represents the velocity at time t, v0 is the initial velocity, a is the acceleration, s(t) is the position at time t, and s0 is the initial position.
a) To find the equation for the height of the ball, s(t), at time t, we use the position equation. Given that the initial height is 50 ft and the initial velocity is 40 ft/sec, the equation becomes:
s(t) = 50 + 40*t - 0.5*20*t^2
= 50 + 40*t - 10*t^2
b) To find the velocity when the ball hits the ground, we need to determine the values of time (t) at which the ball hits the ground. The ball hits the ground when s(t) = 0. We can solve the equation s(t) = 0 to find the value(s) of t:
0 = 50 + 40*t - 10*t^2
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = -10, b = 40, and c = 50:
t = (-40 ± √(40^2 - 4*(-10)*50)) / (2*(-10))
Simplifying further:
t = (-40 ± √(1600 + 2000)) / (-20)
= (-40 ± √(3600)) / (-20)
Since √(3600) = 60, we have:
t = (-40 ± 60) / (-20)
This gives us two possible values for t: t1 = 1 sec and t2 = -5 sec. We discard t2 since we're interested in the positive value when the ball hits the ground. Therefore, the ball hits the ground at t = 1 sec.
To find the velocity at this time, we can use the velocity equation:
v(t) = v0 + a*t
v(1) = 40 + (-20)*1
= 40 - 20
= 20 ft/sec
Therefore, the velocity when the ball hits the ground is 20 ft/sec.
c) To find the time when the ball hits its maximum height, we need to find the vertex of the path. Since the equation for the height is quadratic, we know the vertex occurs at the axis of symmetry, given by:
t = -b / (2a)
In this case, a = -10 and b = 40:
t = -40 / (2*(-10))
= -40 / (-20)
= 2 sec
So, the time when the ball hits its maximum height is 2 seconds.