Explanation:
Let h is the height of the plane above ground. x is the horizontal distance between the ground and the airport. Let s(t) is the distance between the plane and the airport. So,
 ...........(1)
...........(1)
Given, h = 4, x = 40 and s(t) = -20 mph
Differentiate equation (1) wrt t


When x = 40, 



So, the speed of the airplane is 241.14  m/s. Hence, this is the required solution.
 
        
             
        
        
        
Answer:
24m/s²
Explanation:
Given
Distance S = 3m
Time of fall = 0.5sec
Required
Acceleration due to gravity
Using the equation of motion
S = ut+1/2gt²
Substitute the given values
3 = 0+1/2g(0.5)²
3 = 1/2(0.25)g
3 = 0.125g
g = 3/0.125
g = 24
Hence the value for the acceleration of gravity on this new planet is 24m/s²
 
        
             
        
        
        
Answer:
Drawing the triangle:
H / x = tan 52.2 = 1.29
H / (4.6 - x) = tan 28.8 = .550
H = 1.29 x 
H = .55 * 4.6 - .55 x
1.84 x = 2.53        combining equations
x = 1.38 
4.6 - 1.38 = 3.22
Total base of triangle = 1.38 + 3.22 = 4.6
H / x = tan 52,2 = 1.29
H = 1.29 * 1.38 = 1.78 height of triangle
Check: 
1.78 / 3.22 = tan 28.9     
This agrees with the given value of 28.8