Physics, 06.09.2019 03:20 Scourge927

In the olympic shotput event, an athlete throws the shot with an initial speed of 13.0 m/s at a 41.0 ∘ angle from the horizontal. the shot leaves her hand at a height of 1.80 m above the ground.(a) how far does the shot travel? (b) repeat the calculation of the first part for angle 42.5? .(c) repeat the calculation of the first part for angle 45 ? .(d) repeat the calculation of the first part for angle 47.5? .(e) at what angle of release does she throw the farthest?

Answers: 3

Show answers

Answers

Answer from: Aneesa2507

(a) 18.93 m

First of all, we need to find the time it takes for the shot to reach the ground. This can be found by analyzing the vertical motion only, using the equation:

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

where

y(t) is the vertical position at time t

h = 1.80 m is the initial heigth

$u_y = (13.0) sin 41.0^{\circ}=8.53 m/s$ is the initial vertical velocity

t is the time

g = 9.8 m/s^2 is the acceleration of gravity

Substituting y(t)=0, we solve the equation for t, to find the time at which the shot reaches the ground:

0=1.8 + 8.53t -4.9t^2

which gives two solutions: t = -0.19 s and t = 1.93 s. We discard the former solution since it's negative, so the correct time is

t = 1.93 s

Now, the horizontal distance travelled by the shot is given by

x(t) = u_x t

where

$u_x = (13.0) cos 41.0^{\circ}=9.81 m/s$ is the horizontal velocity

t is the time

Substituting t = 1.93 s, we find the range of the shot:

x=(9.81)(1.93)=18.93 m

(b) 18.96 m

We can solve the problem exactly in the same, replacing the previous angle with $42.5^{\circ}$ :

- The initial vertical velocity is

$u_y = (13.0) sin 42.5^{\circ}=8.78 m/s$

So the equation to find the time is

0=1.8 + 8.78t -4.9t^2

which gives t = 1.98 s as solution. The horizontal velocity is

$u_x = (13.0) cos 42.5^{\circ}=9.58 m/s$

And so the range of the shot is

x=u_x t = (9.58)(1.98)=18.97 m

Again, we solve using the new angle $45^{\circ}$ :

- The initial vertical velocity is

$u_y = (13.0) sin 45^{\circ}=9.19 m/s$

So the equation to find the time is

0=1.8 + 9.19t -4.9t^2

which gives t = 2.05 s as solution. The horizontal velocity is

$u_x = (13.0) cos 45^{\circ}=9.19 m/s$

And so the range of the shot is

x=u_x t = (9.19)(2.05)=18.84 m

(d) 18.70 m

Solving using the new angle $47.5^{\circ}$ ,

- The initial vertical velocity is

$u_y = (13.0) sin 47.5^{\circ}=9.58 m/s$

So the equation to find the time is

0=1.8 + 9.58t -4.9t^2

which gives t = 2.13 s as solution. The horizontal velocity is

$u_x = (13.0) cos 47.5^{\circ}=8.78 m/s$

And so the range of the shot is

x=u_x t = (8.78)(2.13)=18.70 m

(e) At $42.5^{\circ}$

We see from the previous parts that the angle that gives the maximum range is $42.5^{\circ}$ . Therefore we can infer an important thing: while for a projectile launched from the ground (h=0) the maximum range is achieved when the angle of launch is $45^{\circ}$ , this is not true for a projectile launched from a height h.

Indeed, it can be demonstrated that for a projectile launched from height h, the angle at which the maximum range is achevied is given by

$\theta = tan^{-1} (\frac{u}{\sqrt{u^2+2gh}})$

And indeed, by substituting u = 13.0 m/s and h = 1.80 m into the formula, we find

$\theta_{max} =42.3^{\circ}$

Which is in agreement with our findings.

Answer from: Quest

32: from largest to smallest, this hierarchy includes eons, eras, periods, epochs, and ages.

33: answer is c

i really hope this answer you out! it makes my day people like you and giving back to the community that has me through school! if you could do me a favor, if this you and this is the very best answer and you understand that all of my answers are legit and top notch. mark as brainliest! and have a awesome day!

Answer from: Quest

another example of kinetic energy would be,

kinetic energy > moving energy

answer > letter choice (c), a hammer falling off a table.

hope that ! : )

Another question on Physics

Physics, 21.06.2019 18:10

Scientists use laser range-finding to measure the distance to the moon with great accuracy. a brief laser pulse is fired at the moon, then the time interval is measured until the "echo" is seen by a telescope. a laser beam spreads out as it travels because it diffracts through a circular exit as it leaves the laser. in order for the reflected light to be bright enough to detect, the laser spot on the moon must be no more than 1.0 km in diameter. staying within this diameter is accomplished by using a special large-diameter laser. part a if λ = 540nm , what is the minimum diameter of the circular opening from which the laser beam emerges? the earth-moon distance is 384,000 km. express your answer to two significant figures and include the appropriate units.

Answers: 1

Answer

Physics, 21.06.2019 22:30

Distinguish between a transverse wave and a longitudinal wave.

Answers: 1

Answer

Physics, 22.06.2019 02:30

You are about to open a container of soy milk but notice that there are instructions to “shake well before serving.” the soy milk is most likely a

Answers: 1

Answer

Physics, 22.06.2019 07:30

Carbon-14 is a radioactive element that undergoes beta decay. which force is responsible for allowing carbon-14 to become stable? electromagnetic gravitational weak nuclear strong nuclear

Answers: 2

Answer

You know the right answer?

In the olympic shotput event, an athlete throws the shot with an initial speed of 13.0 m/s at a 41.0...

Questions