An impact which stops a moving object must do enough work to take away its kinetic energy, so extending the distance moved during the collision reduces the impact force.
Answer:
v₀ = 16.55 m/s
Explanation:
This motion of the ball can be modeled as a projectile motion with following data:
R = Range of Projectile = 27.5 m
θ = Launch Angle = 50°
g = acceleration due to gravity = 9.81 m/s²
v₀ = Initial Speed of Ball = ?
Therefore, using formula for range of projectile, we have:

<u>v₀ = 16.55 m/s</u>
You need to know how much friction that object.
Answer:32 m/s/s
Explanation: since F=M*A, F=16N, M=0.5kg, A= F/M
A=16/0.5
A=32 m/s/s
Answer:
Explanation:
given,
tuning fork vibration = 508 Hz
accelerates = 9.80 m/s²
speed of sound = 343 m/s
observed frequency = 490 Hz


![v_s = v[\dfrac{f_s}{f_o}-1]](https://tex.z-dn.net/?f=v_s%20%3D%20v%5B%5Cdfrac%7Bf_s%7D%7Bf_o%7D-1%5D)
![= 343[\dfrac{508}{490}-1]](https://tex.z-dn.net/?f=%3D%20343%5B%5Cdfrac%7B508%7D%7B490%7D-1%5D)

distance the tunning fork has fallen


=8.1 m
now, time required for the observed will be

now, for the distance calculation


=0.293 m
total distance
= 8.1 + 0.293 = 8.392 m