Answer:
v_y = v_{oy} - g t
where the upward direction is positive, so the arrow represents this speed (blue) must decrease, reach zero and grow in a negative direction as time progresses
Explanation:
In this exercise you are asked to observe the change in velocity in a projectile launch.
If we assume that the friction force is small, the velocity in the x-axis must be constant
vₓ = v₀ₓ
Therefore, the arrow (red) that represents this movement must not change in magnitude.
In the direction of the y axis, the acceleration of gravity is acting, so the magnitude of the velocity in this axis changes
v_y = v_{oy} - g t
where the upward direction is positive, so the arrow represents this speed (blue) must decrease, reach zero and grow in a negative direction as time progresses
True it was thought of as an advantage
The first object is also negatively charged that is why people say opposites attract. have a good day!
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:
initial velocity (u) = 200m/s
final velocity (v) = 350 m/s
time (t) = 15s
acceleration (a) = ?
NOW,
a=v-u/t
a= 350-200/15
a= 50/15
a= 3.3333
Explanation:
it's too easy just u need to understand the question . and go according to it's content .
main thing to memorize is it's simple formula.
I HAVE SOLVE THIS QNA. IN VERY
SIMPLE AND UNDERSTANDABLE FORM.
I høpë u hađ uņdērstøöď ťhìs şølutîóñ
:verý ×wəłł.