Projectile Speed and Distance Calculation

[404]

A projectile is launched with an initial speed of 60.0 m/s at an angle of 30 degrees above the horizontal. It lands on a hillside 4 sec later. Neglect air friction, what is the straight line distance from where the projectile was launched to where it hits the target?

I thought that was surprisingly easy, just use (sin30)*60*4, but I ended up with 207.84m, but the answer book said it was 212m? What did I do wrong?

A soccer player kicks a rock horizontally off a 40.0m high cliff into a pool of water. If the player hears a splash 3 sec later, what is the initial speed given to the rock? Assume speed of sound in air to be 343m/s

I figured out the vertical component of the initial speed to be 28 m/s, but how can you get the horizontal? I think you may be able to get the horizontal distance with the speed of sound?

 

[whozum]

What did I do wrong?

You're looking for the horizontal component of distance.

edit: unless there is more given information, your answer is correct. I think you accidentally wrote sin instead of cos.

I figured out the vertical component of the initial speed to be 28 m/s, but how can you get the horizontal? I think you may be able to get the horizontal distance with the speed of sound?

To solve this one you'll need to realize that the total time (3 sec) is the time it takes for the rock to fall, hit the water, then let the soundwave travel back up to you. Don't worry about horizontal and vertical components, treat it as if he dropped the rock.

 

[Astronuc]

One has to figure the time that the rock falls 40 m. Assume it starts at rest and drops, accelerating with gravity. That is t_f or time falling.

Then the sound take time, t_s to return, and

t_f + t_s = 3 sec

Find the relationship between time and distance in the vertical direction.

The distance the sound traveled = L = \sqrt{{40m}^2+x^2} where x is the horizontal distance traveled.

Assume constant velocity, v_x in x-direction.

 

[404]

Got it. Thank you.