Horizontal speed = 24.0 m/s
height of the cliff = 51.0 m
For the initial vertical speed will are considering the vertical component. Therefore,
Since the student fires the canonical ball at the maximum height of 51 m, the initial vertical velocity will be zero. This means
let's find how long the ball remained in the air.
![\begin{gathered} 0=51-\frac{1}{2}(9.8)t^2 \\ 4.9t^2=51 \\ t^2=\frac{51}{4.9} \\ t^2=10.4081632653 \\ t=\sqrt[]{10.4081632653} \\ t=3.22 \\ t=3.22\text{ s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%200%3D51-%5Cfrac%7B1%7D%7B2%7D%289.8%29t%5E2%20%5C%5C%204.9t%5E2%3D51%20%5C%5C%20t%5E2%3D%5Cfrac%7B51%7D%7B4.9%7D%20%5C%5C%20t%5E2%3D10.4081632653%20%5C%5C%20t%3D%5Csqrt%5B%5D%7B10.4081632653%7D%20%5C%5C%20t%3D3.22%20%5C%5C%20t%3D3.22%5Ctext%7B%20s%7D%20%5Cend%7Bgathered%7D)
Finally, let's find the how far from the base of the building the ball landed(horizontal distance)
Answer:a star’s brightness or luminosity depends on the star’s surface temperature and size.
Explanation:
First of all we know about the ionic bond that the attraction b/w two atoms the atom which lose electrons is become cation and have positive sing which one get electrons it become ionic. so in such sitivation the sodium lose one electons "that's why sodium has 1+ charge in Nacl chemacal bond."
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Explanation:
No surface is perfectly smooth. Everything has irregularities, like bumps and ridges. Static friction is the friction between non-moving objects when their surfaces lock together. When an object is sliding, its surface skips and jumps along the surface of the other object. Because there's less locking between the surfaces, the sliding friction is less than the static friction.
Answer:
Explanation:
a )
the value Vy at the top of the bullet's trajectory will be zero . At the top , only the horizontal component Vx will exist .
b )
At the top Vy = 0
initial velocity will have vertical component = 260 sin30 = 130 m /s
acceleration = - g = - 9.8
v = u - gt
o = 130 - 9.8 t
t = 13.26 s .
c )
Total time = Time of ascent + time of descent
= 13.26 + 13.26 = 26.52 s
d )
Range R = u² sin2Ф / g , where u is initial velocity , Ф is angle of throw , g is acceleration due to gravity .
R = 260² sin 2 x 30 / 9.8
= 5973.8 m .
