A ball is dropped from the roof a building which is 256m high.The height h of the ball at any instant t is described by
h(t)= -16tsqure+256
Find
1)Instantaneous velocity of ball at t= 3sec
2)The time taken by ball to reach the ground level
(not physical education)
1) Instantaneous velocity of the ball at t=3sec can be found by differentiating the height equation, h(t), with respect to time, t.
h(t) = -16t^2 + 256
Taking the derivative of h(t) with respect to t:
h'(t) = -32t
Substituting t=3 into the derivative:
h'(3) = -32 * 3 = -96 m/s
So, the instantaneous velocity of the ball at t=3sec is -96 m/s.
2) To find the time taken by the ball to reach the ground level, we need to set h(t) equal to zero and solve for t.
h(t) = -16t^2 + 256
Setting h(t) equal to zero:
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.
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, and evaluate it at t = 3.
1) Instantaneous velocity at t = 3 seconds:
First, let's find the derivative of the height function with respect to time:
h(t) = -16t^2 + 256
Taking the derivative of h(t) with respect to t:
h'(t) = -32t
Now, substitute t = 3 into the derivative to find the instantaneous velocity:
h'(3) = -32(3)
h'(3) = -96
Therefore, the instantaneous velocity of the ball at t = 3 seconds is -96 m/s.
2) Time taken to reach the ground level:
The ball reaches the ground level when the height is equal to 0. So, we can set the height function equal to 0 and solve for t:
h(t) = -16t^2 + 256 = 0
Subtracting 256 from both sides:
-16t^2 = -256
Dividing both sides by -16:
t^2 = 16
Taking the square root of both sides:
t = ±√16
Since time cannot be negative, the ball would take t = 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 understand the concepts of instantaneous velocity and time taken to reach the ground level in relation to the given equation.
1) Instantaneous Velocity at t = 3sec:
The instantaneous velocity of an object is the derivative of its position function with respect to time. In this case, the position function is h(t) = -16t^2 + 256.
To find the instantaneous velocity at t = 3 sec, we need to find the derivative of h(t) with respect to t. Let's find the derivative:
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 sec is -96 m/s.
2) Time taken by the ball to reach the ground level:
The ball reaches the ground level when its height, h(t), is equal to zero. In this case, we need to solve the equation:
-16t^2 + 256 = 0
Let's solve this equation:
-16t^2 = -256
Divide both sides by -16:
t^2 = 16
Take the square root of both sides:
t = ±4
Since time cannot be negative in this context, 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.