projectile...?

Question

a stuntman on a motorcycle plans to ride at top speed (30m/s) up and off a ramp 2m high inclined @ 45 degrees to the horizontal. calculate the distance travelled from the ramp to the point of impact on the ground.

step up step answers only ppls.

no ressitance or darg...
i need to know the answer for the above answer using suvat equations.

Answer 1

First you have the velocity in the y direction, then use that to find the time in the air. v(in y)= 30sin(45) = 21.21m/s. then solve for t....
y=yo + vo* t+ (1/2)at^2. ...... 0=2 + 21.21t - 4.9t^2 and that gives t= 4.42sec with the quadratic formula.
Using this time you can then solve for the x distance since the x velocity stays constant.
v(in x)= 30cos(45)=21.21
x= v*t = 21.21*4.42= 93.8m

Answer 2

horizontal range =
X = (u^2 sin 2*theta)/g
=(900* sin90) / g
=900/9.8 m
so distance travelled from the ramp
=sqr rt(X^2 + h^2)
where h is the height of the ramp(pythogoras theorem)

Answer 3

first,we resolve the velocity in two components

i) horizontal component (30cos 45 deg.)m/sec
ii)vertical component (30sin45 deg.)m/sec

now, if the upward direction is considered positive for the vertical component,net displacement along the vertical direction is (-2metre)

if,the bike is in motion for t seconds,then

(30sin45 deg.)xt -1/2 x 9.81 x t x t= (--2)

from this eqn. calculate 't'.

so,net horizontal displacement will be

horizontal component of velocity x time
=(30cos45 deg.) x t

Answer 4

Ummm...from front of ramp or end of ramp? How long is the ramp? Drag coefficient? Weight of the projectile?