Projectile Motion of a thrown stone

  • #1

Homework Statement


#1
A student stands at the edge of a cliff and throws a stone
horizontally over the edge with a speed of 18 m/s. The cliff is 50 m above a flat horizontal beach. How long after being released does the stone strike the beach below the cliff ? With what speed and angle of impact does it land ?

-----------------------------------------
#2
A science student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of 10 m/s. The student throws a ball along a path that she judges to make an initial angle of 60*, with the horizontal and to be in line with the track. The student's professor, who is standing on the ground nearby, observes the ball to rise vertically. How high does the ball rise.


2. The attempt at a solution
#1
Vx0 = 18 m/s
d = rt --> x = 18t

Dy = -50 m
a = -9.8 m/s2
-50 = (1/2)(-9.8)t2
which t2 = negative, it is imposible



#2
Vx0 = 10 m/s

Vx0 = 10 (cos 60*) = 5 m/s
Vy0 = 10 (sin 60*) = 8.7 m/s

Am I on the right track on both problem ? Can you guys help me ?
 

Answers and Replies

  • #2
learningphysics
Homework Helper
4,099
6

Homework Statement


#1
A student stands at the edge of a cliff and throws a stone
horizontally over the edge with a speed of 18 m/s. The cliff is 50 m above a flat horizontal beach. How long after being released does the stone strike the beach below the cliff ? With what speed and angle of impact does it land ?

-----------------------------------------
#2
A science student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of 10 m/s. The student throws a ball along a path that she judges to make an initial angle of 60*, with the horizontal and to be in line with the track. The student's professor, who is standing on the ground nearby, observes the ball to rise vertically. How high does the ball rise.


2. The attempt at a solution
#1
Vx0 = 18 m/s
d = rt --> x = 18t

Dy = -50 m
a = -9.8 m/s2
-50 = (1/2)(-9.8)t2
which t2 = negative, it is imposible

t^2 isn't negative. What is t^2?

#2
Vx0 = 10 m/s

Vx0 = 10 (cos 60*) = 5 m/s
Vy0 = 10 (sin 60*) = 8.7 m/s

Am I on the right track on both problem ? Can you guys help me ?

There are two ways to do this... using energy (much easier)... or just using kinematics... have you covered energy yet in your class?
 
  • #3
#1
Oh, It was a mistake !
t = 3.2 s
How can you find the speed and angle of impact does it land ?


#2
I think my teacher haven't covered energy yet !
 
  • #4
learningphysics
Homework Helper
4,099
6
#1
Oh, It was a mistake !
t = 3.2 s
How can you find the speed and angle of impact does it land ?

THe horizontal velocity remains 18m/s. What is the vertical velocity when impact happens?

#2
I think my teacher haven't covered energy yet !

Ok... so suppose the ball is at the bottom of that 60 degree incline... what is the acceleration along the incline? First find the force along the incline, then you can get the acceleration along the incline...
 
  • #5
#1
Vy = -160 m/s


#2
Is the picture below right ?
10 m/s to the East, and then another line 30* N of E,
 
  • #6
learningphysics
Homework Helper
4,099
6
#1
Vy = -160 m/s

Vy should just be 3.2*(-g) = 3.2(-9.8) right?

#2
Is the picture below right ?
10 m/s to the East, and then another line 30* N of E,

Yeah... but I'm a little confused by how the question is asked... is the ball initially going at 10m/s up the 30 degree incline? They say it is in line with the track... does that mean that the ball's horizontal component is 10m/s... ie vcos(30) = 10m/s ?
 
  • #7
#1
how to find the angle ?

#2
I am not sure about it too. They only said the train is running at constant 10 m/s, they didn't said about the velocity of the ball
 
Last edited:
  • #8
learningphysics
Homework Helper
4,099
6
#1
how to find the angle ?

arctan(vy/vx)

#2
I am sure about it too. They only said the train is running at constant 10 m/s, they didn't said about the velocity of the ball

I'm guessing the ball starts at 10m/s... and it's going up an incline at 60degrees... ie E 60degrees N.

So what is the net force acting on the ball along the plane.
 

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