Answer:
a= 92. 13 m/s²
Explanation:
Given that
Amplitude ,A= 0.165 m
The maximum speed ,V(max) = 3.9 m/s
We know that maximum velocity in the SHM given as
V(max) = ω A
ω=Angular speed
A=Amplitude

ω=23.63 rad/s
The maximum acceleration given as
a = ω² A
a= (23.63)² x 0.165 m/s²
a= 92. 13 m/s²
Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².
Answer:
the horizontal distance is 4.355 meters
Explanation:
The computation of the horizontal distance while travelling in the air is shown below:
Data provided in the question is as follows
Velocity = u = 7.70 m/s
H = 1.60 m
R = horizontal direction
Based on the above information
As we know that
R = u × time
where,
Time = 
So,
= 
= 4.355 meters
hence, the horizontal distance is 4.355 meters