Calculating the horizontal distance traveled of an object that was thrown.

In summary, the ball is thrown up onto a roof and lands 3.70s later at a height of 21.0m above the release level. The ball's path just before landing is angled at 54.0˚ with the roof. By using the equation Xf = Xo + 1/2(Vf + Vo)t, the horizontal distance traveled can be calculated to be 33.5m. The original poster had trouble finding the correct answer and requested help. However, it is important for posters to show their own work and not rely on others to simply solve the problem for them.
  • #1
DavidAp
44
0
A ball is thrown up onto a roof, landing 3.70s later at height h = 21.0m above the release level. The ball's path just before landing is angled at θ = 54.0˚ with the roof.
Find the horizontal distance it travels.

Answer: 33.5m


Related Equations:
Xf = Xo + 1/2(Vf + Vo)t
_____________________

My approach on this problem was to find the velocity the moment before impact with the roof and to use basic trigonometry to calculate the horizontal velocity of the ball. From the given facts I can infer that:

Xf = 21m
Xo = 0m
t = 3.70s
Vo = ?
Vf = Vo*sin(54)

Now, plugging this into the equation above I get:
21 = 1/2(Vo*sin(54) + Vo)3.70
42 = (Vo*sin(54) + Vo)3.70
11.35 = .81Vo + Vo
11.35 = 1.81Vo
6.27 = Vo

With Vo I can calculate the horizontal velocity which would be:
Vx = Vo*cos(54)
Vx = 6.27(.59)
Vx = 3.69

And the distance should therefore be:
d = VxT
d = 3.69(3.70)
d = 13.65

However, the answer is not 13.65m and after scratching my head and approaching this problem through multiple, yet similar, angles I finally decided to ask for help. Can somebody help me and point out what I did wrong? I thought I had the right idea but I can't seem to get the right answer?

Thank you for taking the time to read my question.
Sincerely David A.P.
 
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  • #2
dave, can you write the original question AS IT IS. its not clear how the ball is flying. was it thrown vertically upwards or what ? also when it lands on the roof, which angle the [itex]\theta[/itex] is ? a pic will help
 
  • #3
Sure! Here's an image that came with the problem. However, the wording is nearly identical to the problem itself.

14ahqba.jpg
 
  • #4
Concentrate on the vertical motion first. Rather than clutter things with the sine and cosine functions from the outset, just assume that the initial vertical velocity is vyo. Can you solve for it using one of the common kinematic equations from what you're given?
 
  • #5
IssacNewton said:
hi david

here's how i would do it.
[SNIP ALL]

IssacNewton, it is a Physics Forums rule that helpers NOT simply solve homework problems and present complete solutions. Posters must do their own work, with help, hints, and guidance provided as required. Breaking the rule can get one banned from PF.
 

1. How do you calculate the horizontal distance traveled of an object that was thrown?

The horizontal distance traveled of an object that was thrown can be calculated using the equation d = v * t, where d is the distance traveled, v is the initial velocity of the object, and t is the time the object was in flight.

2. What is the initial velocity of an object that was thrown?

The initial velocity, or the velocity at which the object is thrown, can be measured or estimated using a device such as a radar gun or by using the equation v = d/t, where v is the velocity, d is the distance traveled, and t is the time the object was in flight.

3. How do you measure the time an object was in flight when calculating horizontal distance?

The time an object was in flight can be measured using a stopwatch or a device such as a photogate, which measures the time it takes for the object to pass through a specific point. This time can then be used in the equation d = v * t to calculate the horizontal distance traveled.

4. What factors can affect the horizontal distance traveled of a thrown object?

The horizontal distance traveled of a thrown object can be affected by factors such as air resistance, wind speed and direction, and the release angle and height of the object. These factors can cause the object to deviate from a straight trajectory and affect its horizontal distance.

5. Can the horizontal distance traveled of a thrown object be greater than its initial velocity?

Yes, the horizontal distance traveled of a thrown object can be greater than its initial velocity. This is because the object can continue to move forward even after its initial velocity has decreased due to factors such as air resistance. The horizontal distance traveled is also dependent on the release angle and height of the object, which can affect its trajectory and distance traveled.

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