A ball is thrown straight up from the top of the building 120ft y'all with an initial velocity of 72 ft per second. The height s(t) (in feet) of the ball from the ground, at the time (in seconds) is given by s(t)= 120+72t-16t^2. Find the maximum height attained by the ball

The maximum height attained by the ball is 201 feet from the top of the building

Step-by-step explanation:

In this question, we are asked to calculate the maximum height which was reached by a ball which was thrown from the top of a building.

Firstly, we need to understand what happens when a body reaches its maximum height. When a body reaches its maximum height, what happens is that the vertical component of its velocity becomes zero.

Now, we were given a formula for the maximum height here. To get the velocity at that point which is equal to zero, we simply differentiate the equation given ;

Mathematically;

ds(t)/dt = -32t + 72

Let’s equate this to zero to get the time taken to reach the height.

That would be 32t = 72

t = 72/32 = 2.25 seconds

Now to get this height, we simply insert the time taken into the initial maximum height equation before differentiation.

That would be ;

S(2.25) = 120 + 72(2.25) -16(2.25)^2

S = 120 + 162 - 81

S = 201 feet

The maximum height attained by the ball is 201 feet

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step-by-step explanation:

3,6 and 7.

step-by-step explanation:

a) unit place of 37 is 7

so the unit place of cube of 37 will be

7³= 343=3

the answer is 3.

b) unit place of 56 is 6, so the unit place of cube of 56 will be

6³= 216=6

the answer is 6.

c) unit place of 133 is 3, so the unit place of cube of 133 will be

3³= 27= 7.

the answer is 7.