Barry Bonds hits a home run so that the baseball just clears the top row of bleachers, 24.2 m high, located 126 m from the home plate. The ball is hit at an angle of 43° to the horizontal, and air resistance is negligible. Assume that the ball is hit at a height of 1.00 m above the ground. Find the following values.
(a) the initial speed of the ball
m/s
(b) the time at which the ball reaches the cheap seats
s
(c) the velocity of the ball when it passes over the top row
m/s + m/s
(d) the speed of the ball when it passes over the top row
m/s
So I don't get this set up it seems like I just don't have enough info.
for x
Vx= Vcos43
d=126m
t=?
for y
Vyo=Vsin43
Vy= ?
h=23.2 m (b/c the player is on 1 m tall platform)
a= -9.8m/s^2
t= ?
So how can I find V if I don't have t? or Vf?
At first I thought Vf=0 ,but it ask for Vf later...
I don't get it how can I set this up to solve at least one thing?
Can you help me to get started at least?
1 second ago
- 4 days left to answer.
(a) the initial speed of the ball
m/s
(b) the time at which the ball reaches the cheap seats
s
(c) the velocity of the ball when it passes over the top row
m/s + m/s
(d) the speed of the ball when it passes over the top row
m/s
So I don't get this set up it seems like I just don't have enough info.
for x
Vx= Vcos43
d=126m
t=?
for y
Vyo=Vsin43
Vy= ?
h=23.2 m (b/c the player is on 1 m tall platform)
a= -9.8m/s^2
t= ?
So how can I find V if I don't have t? or Vf?
At first I thought Vf=0 ,but it ask for Vf later...
I don't get it how can I set this up to solve at least one thing?
Can you help me to get started at least?
1 second ago
- 4 days left to answer.
-
There's a similar question to this (with answer) in the link. It should help.
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Barry Bonds hits a home run so that the baseball just clears the top row of bleachers, 24.2 m high, located 126 m from the home plate. The ball is hit at an angle of 43° to the horizontal, and air resistance is negligible. Assume that the ball is hit at a height of 1.00 m above the ground. Find the following values.
(a) the initial speed of the ball
(b) the time at which the ball reaches the cheap seats
(c) the velocity of the ball when it passes over the top row
(d) the speed of the ball when it passes over the top row
You are quite right, if the ball was hit at a high initial velocity it could reach a maximum height and then be dropping by the time it passes over the top of the bleachers, so I think you have to assume it was hit at a lower velocity, and has just attained maximum height AT the bleachers. Then all is clear.
Height attained = 23.2 metres.
Initial vertical velocity component = sqrt.(2gh), = sqrt. 19.6 x 23.2, = 21.3m/sec.
Horizontal component = 21.3/(tan 43) = 22.84m/sec.
a) Initial speed = sqrt. (21.3^2 + 22.84^2) = 31.23m/sec.
b) Time to top bleachers = (126m/ 22.84) = 5.52 secs.
c) Velocity at bleachers = 22.84m/sec., horizontally.
d) Speed = 22.84m/sec.
(a) the initial speed of the ball
(b) the time at which the ball reaches the cheap seats
(c) the velocity of the ball when it passes over the top row
(d) the speed of the ball when it passes over the top row
You are quite right, if the ball was hit at a high initial velocity it could reach a maximum height and then be dropping by the time it passes over the top of the bleachers, so I think you have to assume it was hit at a lower velocity, and has just attained maximum height AT the bleachers. Then all is clear.
Height attained = 23.2 metres.
Initial vertical velocity component = sqrt.(2gh), = sqrt. 19.6 x 23.2, = 21.3m/sec.
Horizontal component = 21.3/(tan 43) = 22.84m/sec.
a) Initial speed = sqrt. (21.3^2 + 22.84^2) = 31.23m/sec.
b) Time to top bleachers = (126m/ 22.84) = 5.52 secs.
c) Velocity at bleachers = 22.84m/sec., horizontally.
d) Speed = 22.84m/sec.