Projectile motion - where will the ball land

AI Thread Summary
A boy throws a ball from a 40m high building towards a friend 30m away, with a projection velocity of 20m/s. The trajectory involves calculating the horizontal and vertical components of the throw, which are approximately 16m/s and 12m/s, respectively. To determine how short the ball falls from his friend, the time to reach the ground must be calculated using the appropriate kinematic equations. The initial vertical velocity is zero, and the sign convention for acceleration due to gravity can be chosen as either positive or negative based on the defined direction. Understanding these principles is crucial for accurately solving the projectile motion problem.
physics kiddy
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Homework Statement



A boy standing on the top of a building 40m high throws a ball directly aiming to his friend standing on the ground 30m away from the base of the building. If the projection velocity is 20m/s. Find how short will the ball fall from his friend.

Homework Equations





The Attempt at a Solution



From the picture I have attached, AB = 40 m
BC = 30 m
Using pythagoras theorem, we have AC = 50 m

Using tan θ = p/b we have tan θ = 40/30 so, θ = tan-1(4/3) = 53°

Now, breaking the components,
Horizontally:
20 * cos53° = 16 m/s

Vertically,
20 * sin 53° = 12m/s

My problem begins here :

1) Which formula do I use, s = ut + 1/2 at2 or v = u + at to find the time it takes to reach ground vertically.

2) What's the sign of u and g here.
 

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Use the first equation, s = ut + 1/2 at2 to find the time it takes to reach the ground, u would be 0 since there is no initial vertical velocity.
 
physics kiddy said:

Homework Statement



A boy standing on the top of a building 40m high throws a ball directly aiming to his friend standing on the ground 30m away from the base of the building. If the projection velocity is 20m/s. Find how short will the ball fall from his friend.

Homework Equations





The Attempt at a Solution



From the picture I have attached, AB = 40 m
BC = 30 m
Using pythagoras theorem, we have AC = 50 m

Using tan θ = p/b we have tan θ = 40/30 so, θ = tan-1(4/3) = 53°

Now, breaking the components,
Horizontally:
20 * cos53° = 16 m/s

Vertically,
20 * sin 53° = 12m/s

My problem begins here :

1) Which formula do I use, s = ut + 1/2 at2 or v = u + at to find the time it takes to reach ground vertically.

2) What's the sign of u and g here.

1) You have only intial velocity and you can choose which is appropriate.
2) You can take going up as positive or down as positive. Once you make a convention on the sign of direction, all vectors direction should follow this convention.
 
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