The shot putter should get out of the way before the ball returns to the launch position.
Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.
The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s
t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.
Answer: 0.45 s
Answer:T=1316.21 N
Explanation:
The tension has two components: Vertical and Horizontal. The
horizontal component is ma, the vertical component is mg. Using
Pythagoras theorem, we can find the tension as:
T=((ma)^2 (mg)^2)^(1/2)
So
T=((129*2.84)^2 (129*9.8)^2)^(1/2)
T=1316.21 N
m/s^2 is 39.2266
is the answer If thats what you needed
It would be 12,000 because newton’s third 2nd law states F=ma (force=matter x acceleration) so 30x400 would be your force .
please mark brainliest and i hope this helps!
Answer:
Electric field due to two charges is given as
Explanation:
As we know that two charges are opposite in nature
So the electric field at the mid point of two charges will add together
so the net field is given as
now we have
now we have
