Ok the problem I'm doing has two parts I know how to do part a and got the right answer I think...
I do not know how to do part b
Here's the problem
A student driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp
(a)With what minimum speed msut he dive off the horizontal ramp? The vertical height of the ramp is 1.5 m above the cars and the horizontal distance he must clear is 20 m
ok I got 36 s^-1 m for my answer for part (a)
part (b) is were I need some help
(b) Waht is the new minimum speed if the ramp is now titled upward, so that "take off angle" is 10 degrees above the horizontal, and nothing else is changed?
Let the initial velocity be u, and resolve into vertical (upwards) and horizontal components.
The initial vertical velocity is
u sin(10)
acceleration = -g = -9.81 m/s/s
If the car clears 20m over 1.5 m drop in t seconds, then
-1.5 = u sin(10) -(1/2)g t2
from which
t=sqrt(2*(1.5+u sin(10))/g)
substitute into the horizontal velocity to clear 20 m of cars,
u cos(10) t = 20
u cos(10) sqrt(2*(1.5+u sin(10))/g) =20
Solve the non-linear equation by trial and error or Newton's method to get u=20.13 m/s
a) The time the car is in the air can be found by d = Vi(t) + 0.5(a)(t^2)
Acceleration is 9.81m/s due to gravity. Initial vertical velocity is 0, because the ramp is horizontal. Therefore, the equation becomes d = 0.5(a)(t^2)
d = 0.5(a)(t^2)
-1.5m = 0.5(-9.81m/s/s)(t^2)
t = .553s
The horizontal velocity required can be found by V = d/t
V = d/t
V = 20m/.553s
V = 36m/s
b) The initial vertical velocity is now Vi*sin(10), and the horizontal velocity is Vi*cos(10)
Again, start with d = Vi(t) + 0.5(a)(t^2).
Because of the angle, the equation becomes 1.5 = Vi*sin(10)(t) + 0.5(-9.81)(t^2).
This time, start by solving for t in V = d/t
Because of the angle, the equation becomes V*cos(10) = 20/t
V*cos(10) = 20/t
t = 20/(V*cos(10))
Now, substitute in d/V for t in d = V*sin(10)(t) + 0.5(-9.81)(t^2).
1.5 = V*sin(10)(20/(V*cos(10))) + 0.5(-9.81)(20/(V*cos(10)))^2
Simplify:
1.5 = 20*sin(10)/cos(10) + 0.5(-9.81)*(20/(V*cos(10)))^2
Now solve for V.
1.5 - 3.527 = 0.5(-9.81)*(20/(V*cos(10)))^2
-2.027*V^2 = 0.5(-9.81)*(20/(cos(10)))^2
V = 31.592 m/s
what is newtons method
isn't the formula suppose
to be
y = Vo t - 2^-1 g t^2
?????
-1.5 = u sin(10) -(1/2)g t2
instead of that shouldn't it be
-1.5 = (u sin(10))? -(1/2)g t2
"-1.5 = u sin(10) -(1/2)g t2
from which
t=sqrt(2*(1.5+u sin(10))/g)"
MathMate, I am afraid the equation is d = Vi(t) + 0.5(a)(t^2) from integration from a constant acceleration. I believe you replaced Vi*t with Vi.
Edit: should be a -1.5
-1.5 - 3.527 = 0.5(-9.81)*(20/(V*cos(10)))^2
-5.027*V^2 = 0.5(-9.81)*(20/(cos(10)))^2
V = 20.061 m/s
Indeed, the factor t was missing.
Yes, I have inserted the term t in the equation and obtained 20.0615 m/s., and t is 1.0123 sec.