Answer:
30.63 m
Explanation:
From the question given above, the following data were obtained:
Total time (T) spent by the ball in air = 5 s
Maximum height (h) =.? 
Next, we shall determine the time taken to reach the maximum height. This can be obtained as follow:
Total time (T) spent by the ball in air = 5 s
Time (t) taken to reach the maximum height =.?
T = 2t
5 = 2t
Divide both side by 2
t = 5/2
t = 2.5 s
Thus, the time (t) taken to reach the maximum height is 2.5 s
Finally, we shall determine the maximum height reached by the ball as follow:
Time (t) taken to reach the maximum height = 2.5 s
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) =.? 
h = ½gt²
h = ½ × 9.8 × 2.5²
h = 4.9 × 6.25
h = 30.625 ≈ 30.63 m
Therefore, the maximum height reached by the cannon ball is 30.63 m
 
        
             
        
        
        
Answer:
The new distance is     d = 0.447 d₀
Explanation:
The electric out is given by Coulomb's Law
          F = k q₁ q₂ / r²
This electric force is in balance with tension.
We reduce the charge of sphere B to 1/5 of its initial value ( =q₂ = q₂ / 5) than new distance (d = n d₀)
=q₂ = q₂ / 5) than new distance (d = n d₀)
dat
      q₁ = 
      q₂ = 
      r = d₀
In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points
          F = k q₁ q₂ / d₀²
          F = k q₁ (q₂ / 5) / (n d₀)²
          .k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)
           5 n² = 1
           n = √ 1/5
           n = 0.447
The new distance is
          d = 0.447 d₀
 
        
             
        
        
        
Answer: Your answer is 1250J
Explanation:
K
E
=
1/2
m
v
2
The mass is  
m
=
25
k
g
The velocity is  v
=
10
m
s
−
1
So,
K
E
=
1
/2
x25
x
10
2^2=
1250
J
pls mark brainiest answer  
 
        
             
        
        
        
A) The acceleration is due to gravity at any given point if you look at it vertically, so 

.
b) 

, so 

. We use 

 and then the final speed must be 0 because it stops at the highest point. So 

. Solve for 

 and you get 

c) 

, and then we plug the values: 

 and we already have the time from "b)", so 
![Y_m_a_x = [(32sin(25))*(32sin(25)/10)] - 5(32sin(25)/10)^2](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%20%5B%2832sin%2825%29%29%2A%2832sin%2825%29%2F10%29%5D%20-%205%2832sin%2825%29%2F10%29%5E2)
; then we just rearrange it 
![Y_m_a_x = 10[(32sin(25))^2/100] - 5 [(32sin(25))^2/100]](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%2010%5B%2832sin%2825%29%29%5E2%2F100%5D%20-%205%20%5B%2832sin%2825%29%29%5E2%2F100%5D%20)
 and finally 
![Y_m_a_x = 5[(32sin(25))^2/100] = (32sin(25))^2/20](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%205%5B%2832sin%2825%29%29%5E2%2F100%5D%20%3D%20%2832sin%2825%29%29%5E2%2F20)