Answer:
Option (e)
Explanation:
If a mass attached to a spring is stretched and released, it follows a simple harmonic motion.
In simple harmonic motion, velocity of the mass will be maximum, kinetic energy is maximum and acceleration is 0 at equilibrium position (at 0 position).
At position +A, mass will have the minimum kinetic energy, zero velocity and maximum acceleration.
Therefore, Option (e) will be the answer.
 
        
             
        
        
        
We have that the letter A in the diagram below given as
Amplitude
Option A
<h3>
Amplitude</h3>
Question Parameters:
Amplitude
Crest
Trough
Wavelength
Generally, the amplitude of a wave is the maximum  displacement of the wave in the medium from its initial position. 
Amplitude is denoted with the letter A
Therefore,Amplitude
Option A
For more information on displacement visit
brainly.com/question/989117
 
        
             
        
        
        
consider the motion in x-direction 
 = initial velocity in x-direction = ?
 = initial velocity in x-direction = ?
X = horizontal distance traveled = 100 m 
 = acceleration along x-direction = 0 m/s²
 = acceleration along x-direction = 0 m/s²
t = time of travel = 4.60 sec
Using the equation 
X =  t + (0.5)
 t + (0.5)  t²
 t²
100 =   (4.60)
 (4.60)
  = 21.7 m/s
 = 21.7 m/s 
consider the motion along y-direction 
 = initial velocity in y-direction = ?
 = initial velocity in y-direction = ?
Y = vertical displacement  = 0 m 
 = acceleration along x-direction = - 9.8 m/s²
 = acceleration along x-direction = - 9.8 m/s²
t = time of travel = 4.60 sec
Using the equation 
Y =  t + (0.5)
 t + (0.5)  t²
 t²
0 =  (4.60) + (0.5) (- 9.8) (4.60)²
 (4.60) + (0.5) (- 9.8) (4.60)²
 = 22.54 m/s
 = 22.54 m/s 
initial velocity is given as 
 = sqrt((
 = sqrt(( )² + (
)² + ( )²)
)²) 
 = sqrt((21.7)² + (22.54)²) = 31.3 m/s
 = sqrt((21.7)² + (22.54)²) = 31.3 m/s 
direction: θ = tan⁻¹(22.54/21.7) = 46.12 deg