3- A ball is thrown down vertically with an initial speed of v0 from a height of h.
a)What is its speed just before it strikes the ground?
b)How long does the ball take to reach the ground? What would be the answers to c)part a and
d)part b if the ball were thrown upward from the same height and with the same initial speed? Before solving any equations, decide whether the answers (c) and (d) should be greater than, less than, or the same as in (a) and (b).
Answers
Answer:
this is the answer . hopes it helps u
Explanation:
we will use these equations
First equation of motion eq.1: v=u+at
second equation of motioneq 2: s=ut+
2
1
at
2
third equation of motion eq:3 v
2
=u
2
+2as
here a=−g=−9.8m/s
2
(taking down as the −y direction) for the duration of the motion.
The ground level is taken to correspond to y=0.
(a) With y
0
= h and V
0
, Eq. 3 leads to V=
(−V
0
)
2
−2g(y−y
0
)
=
V
0
2
+2gh
.
The positive root is taken because the problem asks for the speed (the magnitude of the velocity).
(b) We use the quadratic formula to solve Eq. 2 for t, with V
0
replaced with −V
0
.
Δy=−V
0
t−
2
1
gt
2
⇒t=
g
−V
0
+
(
−V
0
)
2
−2gΔy
Where the positive root is chosen to yield t>0.
With y = 0 and y
0
=h, this becomes
t=
g
V
0
2
+2gh−V
0
.
(c) If it were thrown upward with that speed from height h then (in the absence of air friction) it would return to height h with that same downward speed and would therefore yield the same final speed (before hitting the ground) as in part (a).
(d) Having to travel up before it starts its descent certainly requires more time than in part (b). The calculation is quite similar, however, except for now having +V
0
in the equation where we had put in -V
0
in part (b). The details follow: Δy=V
0
t−
2
1
gt
2
⇒t=
g
V
0
+
V
0
2
−2gΔy
with the positive root again chosen to yield t>0. With y=0 and y
0
= h, we obtain
t=
g
V
0
2
−2gh+V
0
.