initial speed of the stuntman is given as
![v = 28 m/s](https://tex.z-dn.net/?f=v%20%3D%2028%20m%2Fs)
angle of inclination is given as
![\theta = 15 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2015%20degree)
now the components of the velocity is given as
![v_x = 28 cos15 = 27.04 m/s](https://tex.z-dn.net/?f=v_x%20%3D%2028%20cos15%20%3D%2027.04%20m%2Fs)
![v_y = 28 sin15 = 7.25 m/s](https://tex.z-dn.net/?f=v_y%20%3D%2028%20sin15%20%3D%207.25%20m%2Fs)
here it is given that the ramp on the far side of the canyon is 25 m lower than the ramp from which she will leave.
So the displacement in vertical direction is given as
![\delta y = -25 m](https://tex.z-dn.net/?f=%5Cdelta%20y%20%3D%20-25%20m)
![\delta y = v_y * t + \frac{1}{2} at^2](https://tex.z-dn.net/?f=%5Cdelta%20y%20%3D%20v_y%20%2A%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20at%5E2)
![-25 = 7.25 * t - \frac{1}{2}*9.8* t^2](https://tex.z-dn.net/?f=-25%20%3D%207.25%20%2A%20t%20-%20%5Cfrac%7B1%7D%7B2%7D%2A9.8%2A%20t%5E2)
by solving above equation we have
![t = 3.12 s](https://tex.z-dn.net/?f=t%20%3D%203.12%20s)
Now in the above interval of time the horizontal distance moved by it is given by
![d_x = v_x * t](https://tex.z-dn.net/?f=d_x%20%3D%20v_x%20%2A%20t)
![d_x = 27.04 * 3.12 = 84.4 m](https://tex.z-dn.net/?f=d_x%20%3D%2027.04%20%2A%203.12%20%3D%2084.4%20m)
since the canyon width is 77 m which is less than the horizontal distance covered by the stuntman so here we can say that stuntman will cross the canyon.
15+3=18km/hour
Think about it like this. The boat is going 15 faster than the river, and the river is going 3 faster than the bank, so the boat is going 18 faster than the river bank
Answer:
t = 2.2 s
Explanation:
Given that,
Height of the roof, h = 24.15 m
The initial velocity of the pumpkin, u = 0
We need to find the time taken for the pumpkin to hit the ground. Let the time be t. Using second equation of kinematics to find it as follows :
![h=ut+\dfrac{1}{2}at^2](https://tex.z-dn.net/?f=h%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2)
Here, u = 0 and a = g
![h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 24.15}{9.8}} \\\\t=2.22\ s](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7B1%7D%7B2%7Dgt%5E2%5C%5C%5C%5Ct%3D%5Csqrt%7B%5Cdfrac%7B2h%7D%7Bg%7D%7D%20%5C%5C%5C%5Ct%3D%5Csqrt%7B%5Cdfrac%7B2%5Ctimes%2024.15%7D%7B9.8%7D%7D%20%5C%5C%5C%5Ct%3D2.22%5C%20s)
So, it will take 2.22 s for the pumpkin to hit the ground.
Answer:
v =2.02
Explanation:
v^2=0.05-4.9
v^2=-4.85
square root both side
v=2.02
^^^^this is a not a perfect square
Answer:
U = -3978.8 J
Explanation:
The work of the gravitational force U just depends of the heigth and is calculated as:
U = -mgh
Where m is the mass, g is the gravitational acceleration and h the alture.
for calculate the alture we will use the following equation:
h = L-Lcos(θ)
Where L is the large of the rope and θ is the angle.
Replacing data:
h = 12.2-12.2cos(58.4)
h = 5.8 m
Finally U is equal to:
U = -70(9.8)(5.8)
U = -3,978.8 J