Aprojectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same level from which it was launched. (a) determine the maximum height reached by the projectile.(b) determine the total time in the air.

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

  sequence  is a continuous and distinctive band of  stars  that appears on plots of stellar  color  versus  brightness. these color-magnitude plots are known as  hertzsprung–russell diagramsafter their co-developers,  ejnar hertzsprungand  henry norris russell. stars on this band are known as  main-sequence stars  or  dwarf stars. these are the most numerous true stars in the universe, and include the earth's  sun.after condensation and ignition of a star, it generates  thermal energy  in its dense core region through  nuclear fusion  of  hydrogen  into  helium. during this stage of the star's lifetime, it is located on the main sequence at a position determined primarily by its mass, but also based upon its chemical composition and age. the cores of main-sequence stars are in  hydrostatic equilibrium, where outward thermal pressure from the hot core is balanced by the inward pressure of  gravitational collapse  from the overlying layers. the strong dependence of the rate of energy generation on temperature and pressure to sustain this balance. energy generated at the core makes its way to the surface and is radiated away at the  photosphere. the energy is carried by either  radiation  or  convection, with the latter occurring in regions with steeper temperature gradients, higher opacity or both.the main sequence is sometimes divided into upper and lower parts, based on the dominant process that a star uses to generate energy. stars below about 1.5 times the  mass of the sun  (1.5  m☉) primarily fuse hydrogen atoms together in a series of stages to form helium, a sequence called the  proton–proton chain. above this mass, in the upper main sequence, the nuclear fusion process mainly uses atoms of  carbon,  nitrogen  and  oxygen  as intermediaries in the  cno cycle  that produces helium from hydrogen atoms. main-sequence stars with more than two solar masses undergo convection in their core regions, which acts to stir up the newly created helium and maintain the proportion of fuel needed for fusion to occur. below this mass, stars have cores that are entirely radiative with convective zones near the surface. with decreasing stellar mass, the proportion of the star forming a convective envelope steadily increases. main-sequence stars below 0.4  m☉undergo convection throughout their mass. when core convection does not occur, a helium-rich core develops surrounded by an outer layer of hydrogen.in general, the more massive a star is, the shorter its lifespan on the main sequence. after the hydrogen fuel at the core has been consumed, the star  evolves  away from the main sequence on the hr diagram, into a  supergiant,  red giant, or directly to a  white dwarf.

The objects are acted upon by only one force. gravity. mass times the acceleration of gravity (-9.81meters/second squared).