Answer:
minimum initial velocity is 21.35 m/s
Explanation:
given data
distance S = 30 m
height h = 30 m
maximum acceleration a = 2 m/s²
to find out
minimum initial velocity that your friend could have thrown the object to enable you to catch
solution
first we get here time with the help of second equation of motion
time = 
..................1
put her value we get
time = 
time = 5.477 second
and that is time which tossed object must be take so we apply here again second equation of motion that is
-S = ut - 0.5 × gt² .......................2
-30 = u× 5.477 - 0.5 ×9.8×5.477²
solve it we get
u = 21.35 m/s
so minimum initial velocity is 21.35 m/s
Answer:
364.4 J
Explanation:
I = Moment of inertia of the forearm = 0.550 kgm²
v = linear velocity of the ball relative to elbow joint = 17.1 m/s
r = distance from the joint = 0.470 m
w = angular velocity
Using the equation
v = r w
17.1 = (0.470) w
w = 36.4 rad/s
Rotational kinetic energy of the forearm is given as
RKE = (0.5) I w²
RKE = (0.5) (0.550) (36.4)²
RKE = 364.4 J
Amplitude, is the answer to the question
Answer:
v= - 27 m/s
Explanation:
Given that
s= t³- 9 t²-27 ( Correct from sources)
As we know that velocity given as
v=3 t ² - 18 t ------------1
As we know that acceleration given as
v=3 t ² - 18 t
a=6 t -18
Given that acceleration is zero (a= 0 )
0 = 6 t - 18
t= 3 sec
Now by putting the values in the equation 1
v=3 t ² - 18 t m/s
v=3 x 3 ² - 18 x 3 m/s
v= 27 - 54 m/s
v= - 27 m/s
