• Support PF! Buy your school textbooks, materials and every day products Here!

Sin contest- Projectile motion

  • Thread starter shane99a
  • Start date
  • #1
1
0
A kid is 5 m from a fence that is 4 m high. He throws a ball at 45° from the horizontal which just grazes the fence. How far beyond the fence does the ball land? You may assume that the ball was thrown from the same level as the ground on the far side of the fence.

Here's how i did it and its probably wrong so i need help:
Vi = 0
θ = 45°
Vf = ?

0.5mv2 = mgh
v = √(2gh)
v = √(19.6*4)
Vf = 8.85 m/s

Vy = 8.85sin45 = 6.26 m/s = Vx

Dy = Vyt - 0.5at2
0 = 6.26t - 4.9t2
t = 1.28s

D = Vxt
D = (6.26)(1.28) = 11 m
Therefore the answer is (11 - 5) 6 m which is the distance the ball travelled beyond the fence.
It looks correct except for the height which I don't know if picking 4 was the right idea and would screw up my entire answer, so please someone help.
 

Answers and Replies

  • #2
674
2
Your initial velocity is not 0 m/s since he originally threw the ball. Also, I wouldn't recommend using conservation of energy for this problem. I recommend using the ballistic equations. For example, there is no reason that your final velocity will be at a 45 degree angle with the horizontal.
 
  • #3
rl.bhat
Homework Helper
4,433
7
Hi shane99a, welcome to PF.
Your calculation of v is wrong. It is not the velocity of the projection of the ball.
It is true only if 4 m is the maximum height. But in the problem it is not mentioned.
So you have to use the equation
Dy = vy*t - 1/2*g*t^2.........(1)
Substitute t = x/vx and solve for vx. In the given problem x is given.
To find maximum range, put Dy = 0 in eq.(1) and solve for x.
 

Related Threads on Sin contest- Projectile motion

  • Last Post
Replies
3
Views
867
Replies
7
Views
1K
Replies
11
Views
2K
Replies
1
Views
1K
Replies
10
Views
1K
Replies
12
Views
2K
  • Last Post
Replies
0
Views
998
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
11
Views
3K
Top