Projectile launch at elevated height on angle

In summary, a baseball player hits a home run with a ball leaving the bat at a velocity of 45m/s at an angle of 36.9° above the horizontal. The ball's height above ground level is initially 1m and then drops to 11m as it clears the fence. Using the quadratic formula and linear motion equations, the time and distance traveled by the ball were calculated to be 5.1 seconds and 198m, respectively. The distance from home plate to the fence at this point can be found by multiplying the average horizontal velocity, 36m/s, by the time, 5.1 seconds, resulting in a distance of approximately 183.6m.
  • #1
rahrahrah1
8
0

Homework Statement



a baseball player hits a home run. The ball's velocity as it leaves the bat is 45m/s at an angle of 36.9° above the horizontal. At the point of impact with the bat, the ball is 1 m above ground level, and as it clears the fence it is back down to 11 m above ground level. How far from home plate is the fence at this point?

Homework Equations


quadratic formula
linear motion equations
basic sin and cos
v=d/t

The Attempt at a Solution



so I started with finding the Vv and the Vh velocties
vh= 45cos6.9= 36.0
Vv= 45sin36.9 = 27.0

then I organized the information I had into two separate headings
Vv
v1: 27 m/s
v2:*27.4m/s
a: -9.8m/s
t:*5.5
d: -1m

and Vh
vave: 36m/s
t:*5.5
d:*198m

*I found the time using the quadratic formula and came up with 5.5 seconds
* I calculated the total distance the ball traveled using the formula d = v×t and got 198 m
* calulated the v2 for the Vv to be 27.4 m/s using linear motion equation
I'm thinking i need a colum for the fence distance now, however, this part is slightly iffy.

Vh for fence
vave:
t:T
d: 198-D

Vv for fence
v1:
v2:
a:-9.8m/s
t:T
d: -11m

At this point I've hit a dead end.
 
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  • #2
The question states the ball is at 11m above the ground

So displacement is 10m.
Calculate your time again.(its not right)

After you have time
You will need to multiply it by v(h)*t to get d.(as you already did)

Thats all.thats your answer

I don't understand what (and why)you are doing in the latter part of your post.
 
  • #3
Hmm, I always manage to overcomplicate an easy question. thanks, I calculated my time to be 5.1 secs.
 
  • #4
Your answer appears correct.
 
  • #5
rahrahrah1 said:
Hmm, I always manage to overcomplicate an easy question. thanks, I calculated my time to be 5.1 secs.


Always draw a diagram :-)
That shortens the problem in so many ways.

Your time is correct :-)
Now all that is left is to find the distance.
 

1. What is the equation for calculating the maximum height of a projectile launched at an angle?

The equation for calculating the maximum height of a projectile launched at an angle is h = (v2sin2θ)/2g, where h is the maximum height, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (9.8 m/s2).

2. How does the launch angle affect the range of a projectile?

The launch angle affects the range of a projectile in that the greater the launch angle, the greater the range. This is because the horizontal component of the initial velocity is directly proportional to the range, and a higher launch angle means a greater initial horizontal velocity.

3. How does the initial velocity affect the time of flight of a projectile?

The initial velocity affects the time of flight of a projectile in that the greater the initial velocity, the longer the time of flight. This is because the horizontal component of the initial velocity remains constant throughout the flight, while the vertical component decreases due to gravity. A higher initial velocity means a longer horizontal distance traveled, resulting in a longer time of flight.

4. What is the optimal launch angle for maximum range of a projectile?

The optimal launch angle for maximum range of a projectile is 45 degrees. This is because at this angle, the initial velocity is divided equally between the horizontal and vertical components, resulting in the greatest range.

5. How does air resistance affect the trajectory of a projectile launched at an angle?

Air resistance, also known as drag, affects the trajectory of a projectile launched at an angle by slowing down the projectile and causing it to deviate from its ideal path. This results in a shorter range and a lower maximum height compared to a projectile launched in a vacuum. The effect of air resistance is greater for projectiles with larger surface areas and lower densities.

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