An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal. a) What is the maximum height reached by the object?
b) What is the total flight time (between launch and touching the ground) of the object?
c) What is the horizontal range (maximum x above ground) of the object?
d) What is the magnitude of the velocity of the object just before it hits the ground?

(a) max. height = 3.641 m

(b) flight time = 1.723 s

(c) horizontal range = 31.235 m

(d) impact velocity = 20 m/s

Above values have been given to third decimal.  Adjust significant figures to suit accuracy required.

Explanation:

This problem requires the use of kinematics equations

v1^2-v0^2=2aS (1)

v1.t + at^2/2 = S (2)

where

v0=initial velocity

v1=final velocity

a=acceleration

S=distance travelled

SI units and degrees will be used throughout

Let

theta = angle of elevation = 25 degrees above horizontal

v=initial velocity at 25 degrees elevation in m/s

a = g = -9.81 = acceleration due to gravity (downwards)

(a) Maximum height

Consider vertical direction,

v0 = v sin(theta) = 8.452 m/s

To find maximum height, we find the distance travelled when vertical velocity = 0, i.e. v1=0,

solve for S in equation (1)

v1^2 - v0^2 = 2aS

S = (v1^2-v0^2)/2g = (0-8.452^2)/(2*(-9.81)) = 3.641 m/s

(b) total flight time

We solve for the time t when the vertical height of the object is AGAIN = 0.

Using equation (2) for vertical direction,

v0*t + at^2/2 = S    substitute values

8.452*t + (-9.81)t^2 = 3.641

Solve for t in the above quadratic equation to get t=0, or t=1.723 s.

So time for the flight = 1.723 s

(c) Horiontal range

We know the horizontal velocity is constant (neglect air resistance) at

vh = v*cos(theta) = 25*cos(25) = 18.126 m/s

Time of flight = 1.723 s

Horizontal range = 18.126 m/s * 1.723 s = 31.235 m

(d) Magnitude of object on hitting ground, Vfinal

By symmetry of the trajectory, Vfinal = v = 20, or

Vfinal = sqrt(v0^2+vh^2) = sqrt(8.452^2+18.126^2) = 20 m/s

emily ruth's annual premium is $1058.50

step-by-step explanation:

emily ruth is a female of 27 years age purchasing 20 year endowment insurance with face value $25000.

from the given table,

for females, the age of a male 3 years younger should be used.

so, we have to check for 24 year male for 20 year endowment.

it is 42.34 for every $1000 face value.

for a face value of $25000,

it is 25 * 42.34 = $1058.50

∴ emily ruth's annual premium is $1058.50

so here is the information for current research:

-nrel's geothermal analysts, scientists, and engineers perform a full range of research from developing advanced methods of locating and accessing geothermal resources to exploring innovative technologies to extract and produce clean energy from the heat of the earth

and here is the information for the technology of geothermal energy:

-it's clean and sustainable

-resources of geothermal energy range from the shallow ground to hot water and hot rock found a few miles beneath the earth's surface, and down even deeper to the extremely high temperatures of molten rock called magma

-almost everywhere, the shallow ground or upper 10 feet of the earth's surface maintains a nearly constant temperature between 50° and 60°f (10° and 16°c)

-geothermal heat pumps can tap into this resource to heat and cool buildings

-a geothermal heat pump system consists of a heat pump, an air delivery system (ductwork), and a heat exchanger-a system of pipes buried in the shallow ground near the building

-in the winter, the heat pump removes heat from the heat exchanger and pumps it into the indoor air delivery system

-in the summer, the process is reversed, and the heat pump moves heat from the indoor air into the heat exchanger

-the heat removed from the indoor air during the summer can also be used to provide a free source of hot water