A ball is dropped from the roof a building which is 256 m high. The height h of the ball at any instant t is described by
h(t) = – 16t2 + 256
Find
i) Instantaneous velocity of ball at t = 3 seconds
ii) The time taken by ball to reach the ground level.
well,
v(t) = -32t
solve for t in
256 - 16t^2 = 0
i) Ah, the good old kinematic equations! To find the instantaneous velocity of the ball at t = 3 seconds, we need to find the derivative of the height function h(t).
So, let's take the derivative of h(t) = -16t^2 + 256:
h'(t) = -32t
Now, substitute t = 3 seconds into the derivative:
h'(3) = -32(3)
h'(3) = -96
Therefore, the instantaneous velocity of the ball at t = 3 seconds is a brisk -96 m/s. That ball is on the fast track to the ground!
ii) To find the time taken by the ball to reach the ground level, we need to find the value of t when the height h(t) is equal to 0. When the ball reaches the ground, its height is 0, right?
So, let's set h(t) = 0 and solve for t:
-16t^2 + 256 = 0
16t^2 = 256
t^2 = 256/16
t^2 = 16
t = ±4
We take the positive root since time cannot be negative, so the ball takes 4 seconds to reach the ground.
So, the clownish conclusion is that the ball took 4 seconds to go from "roof to hoof." Just be glad you're not there to catch it!
To find the instantaneous velocity of the ball at t = 3 seconds, we need to find the derivative of the height function with respect to time.
i) Instantaneous velocity of ball at t = 3 seconds:
The height function h(t) is given by h(t) = -16t^2 + 256.
To find the derivative, we differentiate h(t) with respect to t:
h'(t) = d/dt(-16t^2 + 256)
Differentiating each term separately, we get:
h'(t) = -32t
Substituting t = 3 into h'(t) gives us the instantaneous velocity:
h'(3) = -32(3)
= -96
Therefore, the instantaneous velocity of the ball at t = 3 seconds is -96 units/second.
ii) The time taken by the ball to reach the ground level:
To find the time taken by the ball to reach the ground level, we need to set h(t) = 0 and solve for t.
h(t) = -16t^2 + 256
Setting h(t) = 0:
0 = -16t^2 + 256
Rearranging the equation:
16t^2 = 256
Dividing both sides by 16:
t^2 = 16
Taking the square root of both sides:
t = ±4
Since time cannot be negative, the ball takes 4 seconds to reach the ground level.
Therefore, the time taken by the ball to reach the ground level is 4 seconds.
To find the answers to these questions, we need to use the given equation that describes the height of the ball at any instant in terms of time. Let's solve them step by step.
i) Instantaneous velocity of ball at t = 3 seconds:
To find the instantaneous velocity of the ball at t = 3 seconds, we need to take the derivative of the height function h(t) with respect to time t.
h(t) = -16t^2 + 256
Taking the derivative of h(t) with respect to t, we get:
h'(t) = -32t
Now, substitute t = 3 into the derivative equation:
h'(3) = -32(3) = -96
Therefore, the instantaneous velocity of the ball at t = 3 seconds is -96 m/s.
ii) Time taken by the ball to reach the ground level:
The ground level is when the height, h(t), is equal to 0. So, we need to solve the equation -16t^2 + 256 = 0 to find the time taken by the ball to reach the ground level.
-16t^2 + 256 = 0
Dividing both sides of the equation by -16, we get:
t^2 - 16 = 0
Rearranging the equation:
t^2 = 16
Taking the square root of both sides, we get:
t = ±4
Since we are considering the time t to be positive, the time taken by the ball to reach the ground level is 4 seconds.