Finding Initial Velocity in Projectile Motion Problem

[MorganJ]

I need someone to help me find the answer to this difficult homework problem: A seagull is diving towards a stone beach at an angle of 60 degrees to the vertical releases a clam from a height of 100 meters. The clam hits the beach 3.02 seconds later. To the nearest tenth of a m/s what was its speed when it was released?

I tried vi= m/s/sin(theta), r=2(vi^2/G)sin(60)cos(60)...?help?!?

 

[Chi Meson]

I need someone to help me find the answer to this difficult homework problem: A seagull is diving towards a stone beach at an angle of 60 degrees to the vertical releases a clam from a height of 100 meters. The clam hits the beach 3.02 seconds later. To the nearest tenth of a m/s what was its speed when it was released?

I tried vi= m/s/sin(theta), r=2(vi^2/G)sin(60)cos(60)...?help?!?

I haven't a clue what you think you were doing, but stop.

For this problem, all that matters are vertical components, so think of this as merely a "straight down free-fall" problem.

What is the vertical displacement?
What is the vertical acceleration?
What is the elapsed time?

(No sines or cosines needed so far)

Knowing any three of the five kinematic variables means you can find the other two. Using the three that you know (above) find "initial velocity"; understand that this will actually be the vertical component of the initial velocity.

 

[tomcue]

Hello, I would be happy to help you with this problem. To find the initial velocity of the clam, we can use the equation for projectile motion, which is:

y = y0 + v0t + (1/2)at^2

Where:
y = final position (in this case, 0 meters since the clam hits the beach)
y0 = initial position (100 meters)
v0 = initial velocity (what we are trying to find)
t = time (3.02 seconds)
a = acceleration due to gravity (9.8 m/s^2)

We can rearrange this equation to solve for v0:
v0 = (y-y0)/t - (1/2)at

Plugging in the values given in the problem, we get:
v0 = (0-100)/3.02 - (1/2)(9.8)(3.02)
v0 = -33.11 m/s

However, we need to keep in mind that this is the velocity in the vertical direction. To find the total initial velocity, we need to use the Pythagorean theorem:
vi = √(v0^2 + vx^2)

Since we know that the angle of the dive is 60 degrees, we can use trigonometry to find the horizontal component of the velocity (vx):
vx = v0cos(60)
vx = -33.11cos(60)
vx = -16.55 m/s

Now, we can plug this value into the Pythagorean theorem:
vi = √((-33.11)^2 + (-16.55)^2)
vi = √(1092.54)
vi = 33.05 m/s

Therefore, to the nearest tenth of a m/s, the initial velocity of the clam when it was released was 33.1 m/s. I hope this helps you with your homework problem! If you have any further questions, please let me know.