Kinematics Equations: Solve for Stone's Velocity & Time

[CathyLou]

Hi.

Could someone please help me with the following M1 level Maths question. I would really appreciate any help as I am pretty stuck at the moment.

A stone is catapulted vertically upwards with a velocity of 25 m/s from a point 2 m above the ground. Find (a) its velocity when it hits the ground (b) the time it takes the reach the ground.

I took the upwards direction to be positive.

u = 25

a = -9.8

v = ?

I did not know where to go from there.

Thank you.

Cathy

 

[neutrino]

Do it in two parts. From 2m above the ground to the max. height, then from the max. height to the ground.

 

[CathyLou]

Do it in two parts. From 2m above the ground to the max. height, then from the max. height to the ground.

Thanks for replying.

Could you please tell me which equation to use?

Thank you.

Cathy

 

[neutrino]

Nothing special...it's the usual equations of kinematics.

 

[tony873004]

First, form an intuitive guess to compare your answer too. With an initial velocity of 25 m/s up, it will take roughly 2.5 seconds for it to slow to 0, since gravity will slow it 10 meters per second every second. This is symmetrical on the way down. So 5 seconds later it will be back at the 2 meters above the ground, but traveling 25 meters per second towards the ground. It will only take a fraction of a second to travel the remaining 2 meters to the ground, especially because it already had a 25 m/s head start.

v_f=v_i+0.5at²
y_f=y_i+v_it+0.5at²
y_f=sqrt(y_i+2a delta y)

You should be able to use these formulas and use algebra to re-write them into any form you need. Try to isolate the unknown variable on one side of the equation, while having all known variables on the other side. Or you could re-write it as a 2nd degree polynomial set to 0 and use the quadratic equation, which gives you 2 answers. Sometimes this is useful, and sometimes you just need to discard one of the answers.

 

[CathyLou]

Okay.

Could someone please help me with this one too?

A particle is projected upwards with a speed of 14 m/s. Find for how long it is above 2 m.

I got that:
u = 14
a = -9.8
s = 2
t = ?

Using s = ut + 1/2at^2 I got that 4.9t^2 - 14t + 2 = 0.

Using the quadratic formula I got that t = 2.71 s or 0.15 s (to 3 s. f.) but the answer in the back of the textbook is 2.56 s. Could someone please tell me where I have gone wrong?

Thank you.

Cathy

 

[CathyLou]

First, form an intuitive guess to compare your answer too. With an initial velocity of 25 m/s up, it will take roughly 2.5 seconds for it to slow to 0, since gravity will slow it 10 meters per second every second. This is symmetrical on the way down. So 5 seconds later it will be back at the 2 meters above the ground, but traveling 25 meters per second towards the ground. It will only take a fraction of a second to travel the remaining 2 meters to the ground, especially because it already had a 25 m/s head start.

v_f=v_i+0.5at²
y_f=y_i+v_it+0.5at²
y_f=sqrt(y_i+2a delta y)

You should be able to use these formulas and use algebra to re-write them into any form you need. Try to isolate the unknown variable on one side of the equation, while having all known variables on the other side. Or you could re-write it as a 2nd degree polynomial set to 0 and use the quadratic equation, which gives you 2 answers. Sometimes this is useful, and sometimes you just need to discard one of the answers.

Thanks so much for your help!

Cathy

 

[tony873004]

Okay.

Could someone please help me with this one too?

A particle is projected upwards with a speed of 14 m/s. Find for how long it is above 2 m.

I got that:
u = 14
a = -9.8
s = 2
t = ?

Using s = ut + 1/2at^2 I got that 4.9t^2 - 14t + 2 = 0.

Using the quadratic formula I got that t = 2.71 s or 0.15 s (to 3 s. f.) but the answer in the back of the textbook is 2.56 s. Could someone please tell me where I have gone wrong?

Thank you.

Cathy

You did it right. You just didn't interpret your answers from the quadratic properly. This object will be 2 feet above the ground twice: once on the way up, and once on the way down. 2.71 and 0.15 are the times it will cross the 2 ft mark. It will spend its first 0.15 seconds below 2 feet, and it will remain above 2 feet until 2.71 seconds after launch. After 2.71 seconds it will spend another 0.15 seconds dropping from 2 feet back to the ground. Just subtract your 2 quadratic answers and you'll have the answer from the back of the book. Draw a picture, tracing the path of this object, and mark the 2 foot level. Put 0.15s next to the 2 foot level on the upward part of the trajectory and 2.71s on the 2 foot level on the downward part of your trajectory.

You also seem to be going against standard practice in these problems by the direction of your y-axis. Usually up is regarded as positive. Your quadratic of "4.9t^2 - 14t + 2 = 0." implies that up is negative. You could re-write it as - 4.9t^2, since gravity pulls down, + 14t since your initial velocity is 14 m/s up, -2 since 2 is your initial position minus your final position (0-2=-2). Since you were consistent with up = negative you got the right answer. And it's not wrong to do it this way. Either way is right,but you have to choose one or the other, and most consider up to be positive. So for the sake of consistency, your teacher in your lectures probably considers up to be positive, so you should too. But ask the teacher.