A stone is thrown vertically upwards. On its way it passes point A with a speed v and passes poin B, 30 m higher than A, with speed Find (1) The speed v (ii) The maximum height reached by the stone above point B .
Answers
Answer:
We neglect air resistance, which justifies a= −g = −9.8m/s
2
(taking down as the -y direction) for the duration of the stone’s
motion.
the ball has constant acceleration motion (and we choose y
0
= 0).
(a) We apply third equation of motion to both measurements.
V
B
2
=V
0
2
–2gy
B
⇒(
2
1
V)
2
+ 2g(y
A
+3)=V
0
2
V
A
2
=V
0
2
–2gy
A
⇒V
2
+2gy
A
=V
0
2
We equate the two expressions that each equal V
0
2
and obtain
4
1
V
2
+2gy
A
+2g(3)=V
2
+2gy
A
2g(3)=
4
3
V
2
Which yields V=2
2g
=8.85m/s.
(b) An object moving upward at A with speed V = 8.85 m/s will
reach a maximum height y−y
A
=V
2
/2g=4.00 m above point A (this
is again a consequence of 3rd law of motion, now with the “final” velocity set to zero
to indicate the highest point).
Thus, the top of its motion is 1.00m above point B.