I need to know how to do it and what the answer is

A circus acrobat is shot out of a cannon with an initial upward speed of 34 ft/s. if the acrobat leaves the cannon 4 ft above th

jok3333 [9.3K]

<h2>It will take 0.125 seconds to reach the net.</h2>

Explanation:

Initial speed, u = 34 ft/s = 10.36 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = Final height - Initial height = 8 - 4 = 4 ft = 1.22 m

We have equation of motion, s = ut + 0.5 at²

Substituting

s = ut + 0.5 at²

1.22 = 10.36 x t + 0.5 x -9.81 x t²

4.905t² - 10.36 t + 1.22 = 0

t = 1.99 s or t = 0.125 seconds

Minimum time is 0.125 seconds.

It will take 0.125 seconds to reach the net.

6 0

3 years ago

You do 20 J of work pushing a crate up a ramp. If the output work from the inclined plane is 11 J, then what is the efficiency o

fomenos

Answer:

55%

Explanation:

take efficiency=power output/power input multiply by 100%

5 0

3 years ago

A passenger in a helicopter traveling upwards at 15 m/s accidentally drops a package out the window. If it takes 15 seconds to r

Alexeev081 [22]

Answer:

The helicopter was 1103.63 meters high when the package was dropped.

Explanation:

We consider positive speed as a downward movement

y: height (m)

t: time (s)

v₀: initial speed (m/s)

Δy = v₀t + $\frac{1}{2}$ gt²

Δy= 15 $\frac{m}{s}$ ×15 s + $\frac{1}{2}$ ×9.81 $\frac{m}{s^{2} }$ ×(15 s)²

Δy= 1103.63 m

8 0

3 years ago

A rabbit runs 28 m toward the left in 9 s; then the rabbit suddenly changes direction and runs 18 m toward the right in the next

Viefleur [7K]

Answer:

The answer is 3.111111.

Explanation:

It runs 28 m in the first 9 s, and 28 divided by 9 equals 3.1 and the one goes on forever.

8 0

3 years ago

Vector ????⃗ has a magnitude of 16.6 and is at an angle of 50.5∘ counterclockwise from the +x‑axis. Vector ????⃗ has a magnitude

natka813 [3]

Answer:

For vector u, x component = 10.558 and y component =12.808

unit vector = 0.636 i+ 0.7716 j

For vector v, x component = 23.6316 and y component = -6.464

unit vector = 0.9645 i-0.2638 j

Explanation:

Let the vector u has magnitude 16.6

u makes an angle of 50.5° from x axis

So $u_x=ucos\Theta =16.6\times cos50.5=10.558$

Vertical component $u_y=usin\Theta =16.6\times sin50.5=12.808$

So vector u will be u = 10.558 i+12.808 j

Unit vector $u=\frac{10.558i+12.808j}{\sqrt{10.558^2+12.808^2}}=0.636i+0.7716j$

Now in second case let vector v has a magnitude of 24.5

Making an angle with -15.3° from x axis

So horizontal component $v_x=vcos\Theta =24.5\times cos(-15.3)=23.6316$

Vertical component $v_y=vsin\Theta =24.5\times sin(-15.3)=-6.464$

So vector v will be 23.6316 i - 6.464 j

Unit vector of v $=\frac{23.6316i-6.464}{\sqrt{23.6316^2+6.464^2}}=0.9645i-0.2638j$

8 0

3 years ago