The Student Room Group

Projectile Motion Question #2

A seagull diving towards a stone beach at an angle of 60 degrees to the vertical releases a clam shell from a height of 100 metres. The clam hits the beach 3.0 seconds later. What was the shell's speed when it was released? Take g = 9.81 ms-2.

To find the speed of the shell, you must find the horizontal and vertical components of the shell's velocity when it was released.

For some reason i cant figure out what to actually do. Any hints would be great. Thank you!
Original post by Nirm
A seagull diving towards a stone beach at an angle of 60 degrees to the vertical releases a clam shell from a height of 100 metres. The clam hits the beach 3.0 seconds later. What was the shell's speed when it was released? Take g = 9.81 ms-2.

To find the speed of the shell, you must find the horizontal and vertical components of the shell's velocity when it was released.

For some reason i cant figure out what to actually do. Any hints would be great. Thank you!


Work out the horizontal and vertical components of the Velocity. Also work out the angle using trig
Reply 2
Avatar for Nirm
Nirm
OP
13
Original post by Incubator
Work out the horizontal and vertical components of the Velocity. Also work out the angle using trig

Still stuck :/
Well you can't immediately say anything about the horizontal but you are given a height - which is a vertical distance and an acceleration due to gravity which is also vertical.

so you can start to work out the vertical component of the initial velocity... it's the initial velocity an object would need in order to cover 100m when it was being accelerated at a constant 9.81 ms^-2 for 3 seconds.

s=100
t=3
a=9.81

find u

---
first select a suvat that just contains s,t,a and u (i.e. no v)
Reply 4
Avatar for lerjj
lerjj
13
Original post by Nirm
A seagull diving towards a stone beach at an angle of 60 degrees to the vertical releases a clam shell from a height of 100 metres. The clam hits the beach 3.0 seconds later. What was the shell's speed when it was released? Take g = 9.81 ms-2.

To find the speed of the shell, you must find the horizontal and vertical components of the shell's velocity when it was released.

For some reason i cant figure out what to actually do. Any hints would be great. Thank you!


Focus on finding the vertical velocity, the total should be (relatively) clear if you have a diagram once you've worked this out.

Vertical velocity: you know it falls a distance of 100m in 3.0 seconds whilst accelerating at 9.81m/s^2. Use the equations of motion.
Original post by Nirm
A seagull diving towards a stone beach at an angle of 60 degrees to the vertical releases a clam shell from a height of 100 metres. The clam hits the beach 3.0 seconds later. What was the shell's speed when it was released? Take g = 9.81 ms-2.

To find the speed of the shell, you must find the horizontal and vertical components of the shell's velocity when it was released.

For some reason i cant figure out what to actually do. Any hints would be great. Thank you!


Hi there,

A seagull 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 seconds later. To the nearest tenth of a 2.32 m/s what was its speed when it was released ?

Draw a diagram where the origin is on the ground below the seagull, which is 100 m. high at t=0. Draw a velocity vector of the clam pointing 60 degrees from the vertical pointing down to the right.

I'm going to assume you meant t= 2.32 seconds for how long it takes the clam to hit the ground. Please check the wording of your question and correct my solution if this is not correct.

Using Y = Y0 + (Vy)t + (0.5)At^2 where Y = vertical height of the clam at time t, Y0=starting height of the clam = 100, Vy= vertical starting velocity of the clam = Vsin30, and A = acceleration of gravity = -9.81 m/s^2, V = velocity of seagull and clam at time of release. So
0 = 100 + [Vsin30](2.32) + (0.5)(-9.81)(2.32)^2
V = - 63.4 m/s

So Velocity of seagull and clam at time of release = 63.4 m/s :smile:
(edited 9 years ago)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you