Physics, 29.06.2019 15:20 kaliyab191

Aprojectile is fired up into the air at a speed of 165 m/s at an angle of 75° relative to the horizontal. ignore air drag. a.) determine the maximum height (above where it was thrown) the projectile will reach. b.) determine how long the projectile will be in the air. assume it lands at the same height it was launched. c.)assuming it lands at the same height it was launched at, determine how far the projectile will land from where it was launched. d.) assuming it lands at the same height it was launched at, determine how fast the projectile will be going upon impact. e.) assuming it lands at the same height it was launched at, determine the angle (relative to the horizontal) of impact. °

Answers: 3

Show answers

Answers

Answer from: cammie5292

A) 1296.3 m

The initial velocity of the projectile is

u = 165 m/s

and the angle is

$\theta=75^{\circ}$

So, we can find the initial vertical velocity of the projectile, which is given by

$u_y = u sin \theta = (165 m/s)sin 75^{\circ}=159.4 m/s$

The motion of the projectile along the vertical axis is a uniformly accelerated motion, with constant acceleration

g=-9.8 m/s^2

where the negative sign means the direction is towards the ground. The maximum height is reached when the vertical velocity becomes zero: therefore, we can use the following SUVAT equation

v_y ^2 - u_y^2 = 2gh

where

v_y = 0 is the final vertical velocity

h is the maximum height

Solving for h, we find

$h=-\frac{-u_y^2}{2g}=\frac{-(159.4 m/s)^2}{2(-9.8 m/s^2)}=1296.3 m$

B) 32.5 s

In order to determine how long the projectile will be in the air, we need to find the time t at which the projectile reaches the ground.

We can find it by analyzing the vertical motion only. The vertical position at time t is given by

$y(t) = u_y t + \frac{1}{2}gt^2$

By substituting y(t) = 0, we find the time at which the projectile reaches the ground. We have:

$0= u_y t + \frac{1}{2}gt^2\\0 = t(u_y + \frac{1}{2}gt)$

which has two solutions:

t = 0 --> beginning of the motion

$u_y + \frac{1}{2}gt=0\\t=-\frac{2u_y}{g}=-\frac{2(159.4 m/s)}{-9.8 m/s^2}=32.5 s$

C) 1387.8 m

The range of the projectile can be found by analyzing the horizontal motion only.

In fact, the projectile travels along the horizontal direction by uniform motion, with constant horizontal velocity, given by:

$u_x = u cos \theta = (165 m/s) cos 75^{\circ}=42.7 m/s$

So, the horizontal position at time t is given by

x(t) = u_x t

If we substitute

t = 32.5 s

which is the time at which the projectile reaches the ground, we can find the total horizontal distance covered by the projectile.

So we have:

x= u_x t = (42.7 m/s)(32.5 s)=1387.8 m

D) 165 m/s

The speed of the projectile consists of two independent components:

- The horizontal velocity, which is constant during the motion, and which is equal to

v_x = u_x = 42.7 m/s

- The vertical velocity, which changes during the motion, given by

v_y = u_y + gt

Substituting

u_y = 159.4 m/s

and

t=32.5 s

We find the vertical velocity when the projectile reaches the ground

v_y = 159.4 m/s + (-9.8 m/s^2)(32.5 s)=-159.4 m/s

which is the same as the initial vertical velocity, but with opposite direction.

Now that we have the two components, we can calculate the actual speed of the projectile:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(42.7 m/s)^2+(-159.4 m/s)^2}=165 m/s$

and the final speed is exactly equal to the initial speed, since according to the conservation of energy, the projectile has lost no energy during the motion.

E) $-75^{\circ}$

The angle of impact is given by

$\theta = tan^{-1} (\frac{|v_y|}{v_x})$

where

|v_y| = 159.4 m/s is the final vertical velocity

v_x = 42.7 m/s is the final horizontal velocity

We have taken the absolute value of v_y since v_y is negative; this means that the resulting angle will be BELOW the horizontal. So we have:

$\theta = tan^{-1} (\frac{159.4 m/s}{42.7 m/s})=75^{\circ}$

which means $-75.0^{\circ}$ , below the horizontal.

Answer from: Quest

50 inches should be the resultant vector

Answer from: Quest

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow

Other questions on the subject: Physics

Physics, 21.06.2019 22:00, homework2358

There is a theory that indicates that dinosaurs became extinct when about 65 million years ago, a large asteroid hit the earth surface. dust caused by this collision blocked the sunlight reaching the earth's surface and many forms of life became extinct due to the cold. fearing this threat, how large the radius of an asteroid should you be looking for if the dangerous asteroid size is approximately the same as the one that killed the dinosaurs? available data suggests that about 18% of that asteroid's mass ended up as a dust spread evenly over earth after eventually settling out of the upper atmosphere. about 0.0180 g/cm^2 of dust, which is chemically different than the earth's rock, covered the earth's surface. typical asteroids have a density of about 1.9 g/cm^3. now that we know the size of the asteroid, how much energy was released during impact, assuming all of it was just the kinetic energy of the asteroid right before the impact?

Answers: 1

continue

Physics, 22.06.2019 11:20, leandrogarin37p2g5ds

Suppose a diode consists of a cylindrical cathode with a radius of 6.200×10^−2 cm , mounted coaxially within a cylindrical anode with a radius of 0.5580 cm . the potential difference between the anode and cathode is 260 v . an electron leaves the surface of the cathode with zero initial speed (v initial=0). find its speed vfinal when it strikes the anode.

Answers: 1

continue

Physics, 22.06.2019 15:50, bapefer498

Ryan is examining the energy of the particles in a bar of gold. what is ryan most likely studying?

Answers: 2

continue

Physics, 22.06.2019 18:00, maren17

Air enters a gas turbine with two stages of compression and two stages of expansion at 100 kpa and 17°c. this system uses a regenerator as well as reheating and intercooling – the intercooler returns the air to the inlet temperature. the pressure ratio across each compressor is 4 ; 300 kj/kg of heat are added to the air in each combustion chamber; and the regenerator operates perfectly while increasing the temperature of the cold air by 20°c. determine the system’s thermal efficiency. assume isentropic operations for all compressor and the turbine stages and use constant specific heats at room temperature. (0.378)

Answers: 3

continue

You know the right answer?

Aprojectile is fired up into the air at a speed of 165 m/s at an angle of 75° relative to the horizo...