Solve Projectile Problem: Calculate Impact Speed & Angle

[Mo]

Ack! I thought i knew this!

"A car manufacturer test crash resistance by driving test vehicles off a horiztontal platform so that they fall to a concrete surface below.If this car is driven off at 20m/s and the platform is 15m above the ground calculate the impact angle and speed"

Im not to fussed about the impact angle just yet, i tried to work the speed out...

u = 0
a = 9.8
s (displacement) = 15
v = ?
t = ?

For this i need to use the equation: $$v^2 = u^2 + 2as$$

$$v^2 = 2 \times 9.8 \times 15$$

$$v^2 =294$$

$$v = 17.15 m/s$$



But the answer in the book is different, it says 26.3 m/s

Im most probably making an incredibly stupid mistake, please spot it!

thanks,

Regards,
Mo

 

[da_willem]

I'd try using the standard equations for projectile motion for constant accelertions:

$$x(t)=x(0) + v_x (0) t +\frac{1}{2}a_x t^2$$
$$y(t)=y(0) + v_y (0) t +\frac{1}{2}a_y t^2$$

Just find the right parameters and solve...

 

[Mo]

Ahh! I've never come across those before!

 

[da_willem]

Well it gives you the motion (horizontal coordinate x and vertical coordinate y) as a function of time t. The first term is the initial position at t=0. The second term says the coordinates change in time proportional to the initial velocity (assuming no friction). The last term tells you how the coordinates change under a constant acceleration a.

I'm sure you understand the first two terms, and for this problem there is an acceleration (gravity!) involved so you will have to know how to use the last term or use conservation of energy.

 

[Doc Al]

For this i need to use the equation: $$v^2 = u^2 + 2as$$

$$v^2 = 2 \times 9.8 \times 15$$

$$v^2 =294$$

$$v = 17.15 m/s$$

This gives you the vertical component of the velocity. Now add the horizontal component. (Don't forget that they are perpendicular.)

 

[vincentchan]

you have already found the y component of final speed...now you need the x component

 

[Mo]

Thank you very much Doc Al! . I am still a bit confused as to why we have to combine the two components , as they are completely independent of each other. But I've got the answer now and ill have to read up on it.

Thanks again

Regards,
Mo