<h2>
After 26.28 seconds projectile returns 26.28 seconds.</h2>
Explanation:
Initial velocity = 450 ft/s = 137.16 m/s
Angle, θ = 70°
Consider the vertical motion of projectile,
When the projectile return to the ground we have
Displacement, s = 0 m
Acceleration, a = -9.81 m/s²
Initial velocity, u = 137.16 x sin70 = 128.89 m/s
Substituting in s = ut + 0.5 at²
s = ut + 0.5 at²
0 = 128.89 x t + 0.5 x (-9.81) x t²
t² - 26.28 t = 0
t ( t- 26.28) = 0
t = 0 s or t = 26.28 s
After 26.28 seconds projectile returns 26.28 seconds.
Answer:
Vx= 11.0865(m/s)
Vy= 6.4008(m/s)
Explanation:
Taking into account that 1m is equal to 0.3048 ft, the takeoff speed in m / s will be:
V= 42(ft/s) × 0.3048(m/ft) = 12.8016(m/s)
The take-off angle is equal to 30 °, taking into account the Pythagorean theorem the velocity on the X axis will be:
Vx= 12.8016 (m/s) × cos(30°)= 11.0865(m/s)
And for the same theorem the speed on the Y axis will be:
Vy= 12.8016 (m/s) × sen(30°)= 6.4008(m/s)
Answer:
a

b
The value is 
Explanation:
From the question we are told that
The mass is
The spring constant is 
The instantaneous speed is 
The position consider is x = 0.750A meters from equilibrium point
Generally from the law of energy conservation we have that
The kinetic energy induced by the hammer = The energy stored in the spring
So

Here a is the amplitude of the subsequent oscillations
=> 
=> 
=> 
Generally from the law of energy conservation we have that
The kinetic energy by the hammer = The energy stored in the spring at the point considered + The kinetic energy at the considered point

=> 
=> 