Solve Projectile Problem: Calculate Impact Speed & Angle

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A car manufacturer tests crash resistance by driving vehicles off a horizontal platform at 20 m/s from a height of 15 m. The initial calculations for impact speed using the equation v^2 = u^2 + 2as resulted in 17.15 m/s, which differs from the book's answer of 26.3 m/s. The discussion emphasizes the need to consider both vertical and horizontal components of velocity, as they are independent but must be combined to find the total impact speed. The confusion arises from understanding how to apply projectile motion equations correctly. Ultimately, the correct approach involves using both components to calculate the overall impact speed accurately.
Mo
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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
 
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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...
 
Ahh! I've never come across those before!
 
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.
 
Mo said:
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.)
 
you have already found the y component of final speed...now you need the x component
 
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
 
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