When the velocity v of an object is very large, the magnitude of the force due to air resistance is proportional to v squared with the force acting in opposition to the motion of the object. A shell of mass 2 kg is shot upward from the ground with an initial velocity of 600 m/sec. If the magnitude of the force due to air resistance is (0.1)v squared, when will the shell reach its maximum height above the ground? What is the maximum height? Assume the acceleration due to gravity to be 9.81 m divided by s squared.

Step-by-step explanation:

The model fo the shell is given by the following equation of equilibrium:

This first-order differential equation has separable variables, which are cleared herein:

The solution of this integral is:

3.4 pizzas

step-by-step explanation:

given is a scatter plot showing number of pizzas sold during weeks for different issues of coupon

y represents number of pizzas sold and x number of coupons issued

y is marked verticaly and x horizontally

the trend line is given as

y =3.4x+43

this says that even when no coupon is issued pizzas sold will be 43

and rate of change of pizza sold for 1 unit of coupon = slope = 3.4

i.e. when x increases by 1 unit y increases by 3.4 units

hence for each additional coupon issued 3.4 pizzas more will be sold