An archer puts a 0.30-kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m. Assuming th
at all the energy goes into the arrow, with what speed does the arrow leave the bow?
2 answers:
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Answer:
Explanation:
Let the length of the string is L.
Let T be the tension in the string.
Resolve the components of T.
As the charge q is in equilibrium.
T Sinθ = Fe ..... (1)
T Cosθ = mg .......(2)
Divide equation (1) by equation (2), we get
tan θ = Fe / mg
As θ is very small, so tanθ and Sinθ is equal to θ.
Answer: Part(a)=0.041 secs, Part(b)=0.041 secs
Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction
now we know that
a=-9.81 m/s^2 ( negative because it is pulling the player downwards)
we also know that
s=76 cm= 0.76 m ( maximum s)
using kinetic equation
where v is final velocity which is zero at max height and u is it initial
hence
now we can find time in the 15 cm ascent
using quadratic formula
t=0.0409 sec
the answer for the part b will be the same
To find the answer for the part b we can find the velocity at 15 cm height similarly using
where s=0.76-0.15
as the player has traveled the above distance to reach 15cm to the bottom
when the player reaches the bottom it has the same velocity with which it started which is 3.861
hence the time required to reach the bottom 15cm is
t=0.0409