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AS Level Projectiles

I am having some difficulty with understanding projectiles and would greatly appreciate if someone could explain to me how I solve a question such as the one posted below. Help is much appreciated.

A stunt driver has to cross a gap 20M wide. If the angle of of the ramp is 45°, what must his minimum take off speed be in Km/h?

Reply 1

sig.fig.

Are we to assume that the landing surface is elevated to the same height as the edge of the ramp?

Reply 2

rahmed

i remember AS physics, quite boring i believe - LOOONG

Reply 3

AlphaNumeric

rahmed

i remember AS physics, quite boring i believe - LOOONG

Then why post? True, some parts of AS Physics are not particularly riveting but it does lead to some interesting things.

The projectile equation relevant to this is

R = \frac{V^{2}}{g}\sin 2\theta

where R is the range and V is the velocity. Here

\theta = 45

R = \frac{V^{2}}{g}\sin 90 = \frac{V^{2}}{g}

R = 20 and g = 9.8 so you get

20 = \frac{V^{2}}{9.8}

V^{2} = 20 \times 9.8 = 196

V = 14m/s

1m/s = 3.6km/h, so 14m/s = 50.4km/h

Reply 4

steve2005

This should help you check your answer.
http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra9

Reply 5

john !!

if you know calculus, the man's acceleration vector is due to his weight, (using i is across the bridge and j is upwards as basis vectors)

a = -gj

integrate, knowing the constant of integration is the initial velocity, v = (|v|/√2)(i + j)

v = (|v|/√2)i + (|v|/√2 - gt)j

integrate once more, with the constant this time being 0 as the initial displacement is zero,

r = (t|v|/√2)i + (t|v|/√2 - gt²/2)j

you want to know the time when the stuntman reaches the ground on the other side, so put the vertical component of the position vector equal to 0:

(t|v|/√2 - gt²/2) = 0
|v|/√2 = gt/2
|v|√2/g = t

and plug this into the equation of the horizontal velocity, putting the total horizontal velocity equal to 20m:

(|v|√2/g)(|v|/√2) = 20
v²/g = 20
v = 14 ms^-2

which (thankfully) matches AlphaNumeric's number