David kicks a soccer ball off the ground and in the air, with an initial velocity of 35 feet per second. Using the formula H(t) = −16t2 + vt + s, what is the maximum height the soccer ball reaches?

Step-by-step explanation:

The function for this problem is: h(t) = -16(t)^2 + vt + s h= the height t= time v= velocity s= starting height With the information given, we know that the starting height is 0, since it was from the ground, and the velocity of the ball is 35 feet per second. Inserting the these information into the equation, we get: h(t) = -16(t)^2 + 35t Now the question asks to find the maximum height. It can be done by using a grapher to graph the maximum of the parabola. It could also be done by finding the vertex, which would be the maximum, of the graph by using x= -b/(2a), where b is equal to 35 and a is equal to -16. We get x=35/32, the x-value of where the vertex lies. You can use this value as the t-value in the previous equation to find the h-value of the vertex. When you do, you get h= 19.1 feet, or answer D.

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2.56 is the width of the garden

step-by-step explanation:

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v