if a ball is thrown vertically upward with an initial velocity of 48 ft/s, then its height, s, after t seconds is given by s(t)= 48t-16t^2.
a) what is the velocity of the ball after 1 second?
b) when does the ball hit the ground?
c) with what velocity does the ball hit the ground?
a) evaluate using t=1
b) solve for t when s=0
c) using t from (b) evaluate s'(t)
since t = 2,
i) at 0.5 seconds,
y= (52(2.5)-16(2)^2)/0.5=-13.6
ii) (52(2.1)-16(2)^2)/0.1=-12.8
and just add the time to 2 and divide by time
a) To find the velocity of the ball after 1 second, we need to find the derivative of the height function with respect to time.
The height function is given by s(t) = 48t - 16t^2.
Taking the derivative of s(t) with respect to t, we get:
s'(t) = 48 - 32t.
Substituting t = 1 into the derivative, we find:
s'(1) = 48 - 32(1) = 48 - 32 = 16 ft/s.
Therefore, the velocity of the ball after 1 second is 16 ft/s.
b) To find when the ball hits the ground, we need to find the value of t when the height of the ball, s(t), is equal to zero.
The height function is given by s(t) = 48t - 16t^2.
Setting s(t) = 0, we have:
48t - 16t^2 = 0.
Factoring out t, we get:
t(48 - 16t) = 0.
This equation is satisfied when t = 0 or 48 - 16t = 0.
Solving 48 - 16t = 0, we find:
-16t = -48,
t = 3.
Therefore, the ball hits the ground after 3 seconds.
c) To find the velocity at which the ball hits the ground, we can use the derivative of the height function.
The derivative of s(t) is given by s'(t) = 48 - 32t.
Substituting t = 3 into the derivative, we find:
s'(3) = 48 - 32(3) = 48 - 96 = -48 ft/s.
Therefore, the ball hits the ground with a velocity of -48 ft/s. Note that the negative sign indicates that the ball is moving downward.
To answer these questions, we will use the given equation for the height of the ball, which is s(t) = 48t - 16t^2. We can differentiate this equation to find the velocity and determine when the ball hits the ground.
a) To find the velocity of the ball after 1 second, we need to find the derivative of the height function with respect to time, t. The derivative represents the rate of change of the height, which is equivalent to the velocity.
Let's find the derivative of the height function s(t) = 48t - 16t^2:
s'(t) = 48 - 32t
Now, substitute t = 1 into s'(t) to find the velocity after 1 second:
s'(1) = 48 - 32(1)
s'(1) = 48 - 32
s'(1) = 16
Therefore, the velocity of the ball after 1 second is 16 ft/s.
b) To determine when the ball hits the ground, we need to find the value of t when the height, s(t), becomes zero. This means the ball will hit the ground at the time when the height function equals zero.
Let's set the height function s(t) = 0 and solve for t:
0 = 48t - 16t^2
Now let's factor out common terms:
0 = 16t(3 - t)
From this equation, we can see that either t = 0 or (3 - t) = 0. Solving for t in each case:
For t = 0, we have:
0 = 16(0)(3 - 0)
0 = 0
For 3 - t = 0, we have:
t = 3
Therefore, the ball hits the ground at t = 0 seconds and t = 3 seconds.
c) To find the velocity with which the ball hits the ground, we need to find the velocity at t = 3 seconds. We can substitute this value of t into the derivative of the height function, s'(t).
Let's find the derivative of s(t) with respect to t again:
s'(t) = 48 - 32t
Now, substitute t = 3 into s'(t) to find the velocity at t = 3 seconds:
s'(3) = 48 - 32(3)
s'(3) = 48 - 96
s'(3) = -48
Therefore, the ball hits the ground with a velocity of -48 ft/s. The negative sign indicates that the velocity is downwards.