Answer:
We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at the release point (the initial position for the ball as it begins projectile motion in the sense of §4-5), and we let θ
0
be the angle of throw (shown in the figure). Since the horizontal component of the velocity of the ball is v
x
=v
0
cos40.0°, the time it takes for the ball to hit the wall is
t=
v
x
Δx
=
(25.0m/s)cos40.0
o
22.0m
=1.15s.
(a) The vertical distance is
Δy=(v
0
sinθ
0
)t−
2
1
gt
2
=(25.0m/s)sin40.0
o
(1.15s)−
2
1
(9.80m/s
2
)(1.15s)
2
=12.0m.
(b) The horizontal component of the velocity when it strikes the wall does not change from its initial value: v
x
=v
0
cos40.0°=19.2m/s.
(c) The vertical component becomes (using Eq. 4-23)
v
y
=v
0
sinθ
0
−gt=(25.0m/s)sin40.0
o
−(9.80m/s
2
)(1.15s)=4.80m/s.
(d) Since v
y
>0 when the ball hits the wall, it has not reached the highest point yet.
Answer:
horizontal component of the velocity of the ball is vx=v0cos40.0°, the time it takes for the ball to hit the wall is
t=Δx/vx=(25.0m/s)cos40.0o22.0m=1.15s.
(a) The vertical distance is
Δy=(v0sinθ0)t−21gt2=(25.0m/s)sin40.0o(1.15s)−21(9.80m/s2)(1.15s)2=12.0m.
(b) The horizontal component of the velocity when it strikes the wall does not change from its initial value: vx=v0cos40.0°=19.2m/s.
(c) The vertical component becomes (using Eq. 4-23)
vy=v0sinθ0−gt=(25.0m/s)sin40.0o−(9.80m/s2)(1.15s)=4.80m/s.
(d) Since vy>0 when the ball hits the wall, it has not reached the highest point yet.
Explanation:
