Given :
Two forces act on a 6.00-kg object. One of the forces is 10.0 N.
Acceleration of object 2 m/s².
To Find :
The greatest possible magnitude of the other force.\
Solution :
Let, other force is f.
So, net force, F = 10 + f.
Now, acceleration is given by :
Therefore, the greatest possible magnitude of the other force is 2 N.
Hence, this is the required solution.
Answer:
(a) - 165.032 m/s
(b) 238.37 m/s
Explanation:
initial horizontal velocity, ux = 172 m/s
height, h = 1390 m
g = 9.8 m/s^2
Let it strikes the ground after time t.
Use second equation of motion in vertical direction
-1390 = 0 - 0.5 x 9.8 x t^2
t = 16.84 second
(a) Let vy be the vertical component of velocity as it strikes the ground
Use first equation of motion in vertical direction
vy = uy - gt
vy = 0 - 9.8 x 16.84
vy = - 165.032 m/s
Thus, the vertical component of velocity as it strikes the ground is 165.032 m/s downward direction.
(b)
The horizontal component of velocity remains constant throughout the motion.
vx = 172 m/s
vy = - 165.032 m/s
The resultant velocity is v.
v = 238.37 m/s
Thus, teh velocity with which it hits the ground is 238.37 m/s.