Answer:
Vf = 41.6 [m/s]
Explanation:
To solve this problem we must use the equations of kinematics.
Vf² = Vo² + (2*g*y)
where:
Vf = final velocity [m/s]
Vo = initial velocity = 0
g = gravity acceleration = 9.81 [m/s²]
y = height = 88.2 [m]
Note: The positive sign of the equation tells us that the acceleration of gravity goes in the direction of motion.
Vf² = Vo² + (2*g*y)
Vf² = 0 + (2*9.81*88.2)
Vf = (1730.48)^0.5
Vf = 41.6 [m/s]
Answer:
We could get the time taken by the ball to return back to earth, using the formula:
s = u t + ½ a t², where
s = displacement of the body moving with initial velocity u, acceleration 'a' in time t.
In the present case s=0 (as the ball returns back to starting time)
u= 30 m/s; a = -10 m/s² ( negative sign as a is in opposite direction to u); t=?
0 = 30 t - ½ ×10 ×t²; ==> 5 t = 30, t= 6 second.
So ball will return back after 6 second after being thrown up.
Explanation:
I looked it up
Hope this helps
To solve this problem we will use the theorems related to the conservation of momentum to determine the final speed of the system, once the bullet hits it. From there, we will calculate with the kinematic equations of linear motion, the height traveled with that speed. The conservation of momentum say us that,
Where,
mass of bullet
mass of block of wood
Initital velocity of bullet and block of wood respectively.
Final velocity
Our values are given as,
Replacing we have that
Note that the initial velocity of wood block is 0 because is at rest
Now using the kinematic equation of motion we have that
As 
Therefore the height reached by block after bullet embedding in it is 0.49 m
