Can someone explain please ???

A stuntwoman is going to attempt a jump across a canyon that is 77 m wide. The ramp on the far side of the canyon is 25 m lower

charle [14.2K]

Answer:

She will make the jump.

Explanation:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

First we will consider horizontal motion of stunt women

Displacement = 77 m, Initial velocity = 28 cos 15 = 27.05 m/s, acceleration = 0

Substituting

$77= 27.05t+\frac{1}{2} *0*t^2\\ \\ t=77/27.05=2.85 seconds$

So she will cover 77 m in 2.85 seconds

Now considering vertical motion, up direction as positive

Initial velocity = 28 sin 15 = 7.25 m/s, acceleration =acceleration due to gravity = -9.8 $m/s^2$ , time = 2.85

Substituting

$s=7.25*2.85-\frac{1}{2}*9.8*2.85^2=20.69-39.80 =-10.11 m$

So at time 2.85 stunt women is 10.11 m below from starting position, far side is 25 m lower. So she will be at higher position.

So she will make the jump.

6 0

3 years ago

1. A man throws a ball up with a velocity of 30 m/s. How high does it get?

ElenaW [278]

Yes the answer is g=5

8 0

3 years ago

A В C D At which point will an image be formed? ОА Ов Ос OD

Anettt [7]

Answer:

oa

Explanation:

it may be oa is the right answer for this question

but I don't know properly

8 0

2 years ago

A cannon ball is fired directly upward with a velocity of 160 m/s. How long does it take to fall back to the ground? s How fast

Andrej [43]

To answer this problem, we will use the equations of motions.

Part (a):
For the ball to start falling back to the ground, it has to reach its highest position where its final velocity will be zero.
The equation that we will use here is:
v = u + at where
v is the final velocity = 0 m/sec
u is the initial velocity = 160 m/sec
a is acceleration due to gravity = -9.8 m/sec^2 (the negative sign is because the ball is moving upwards, thus, its moving against gravity)
t is the time that we want to find.
Substitute in the equation to get the time as follows:
v = u + at
0 = 160 - 9.8t
9.8t = 160
t = 160/9.8 = 16.3265 sec
Therefore, the ball would take 16.3265 seconds before it starts falling back to the ground

Part (b):
First, we will get the total distance traveled by the ball as follows:
s = 0.5 (u+v)*t
s = 0.5(160+0)*16.3265
s = 1306.12 meters
The equation that we will use to solve this part is:
v^2 = u^2 + 2as where
v is the final velocity we want to calculate
u is the initial velocity of falling = 0 m/sec (ball starting falling when it reached the highest position, So, the final velocity in part a became the initial velocity here)
a is acceleration due to gravity = 9.8 m/sec^2 (positive as ball is moving downwards)
s is the distance covered = 1306.12 meters
Substitute in the above equation to get the final velocity as follows:
v^2 = u^2 + 2as
v^2 = (0)^2 + 2(9.8)(1306.12)
v^2 = 25599.952 m^2/sec^2
v = 159.99985 m/sec
Therefore, the velocity of the ball would be 159.99985 m/sec when it hits the ground.

6 0

3 years ago

A car tire rotates with an average angular speed of 32 rad/s. In what time interval will the tire rotate 3.5 times? Answer in un

stiks02 [169]

Answer:

$\Delta t = 0.687\ s$

Explanation:

given,

Angular speed of the tire = 32 rad/s

Displacement of the wheel = 3.5 rev

Δ θ = 3.5 x 2 π

= 7 π rad

now,

Time interval of the car to rotate 7π rad

using equation

$\omega_{avg} =\dfrac{\Delta \theta}{\Delta t}$

$\Delta t=\dfrac{7\pi}{32}$

$\Delta t = 0.687\ s$

Time taken to rotate 3.5 times is equal to 0.687 s.

3 0

3 years ago