30 Ramp Platform
Check out this page if you are looking for 30 Ramp Platform
US $4,000.00
first year university physics problem, help?
I have a physics exam coming up, and i have no idea how to do this type of problem. thanks to anyone who is able to help. i will choose Best Answer.
A toy car is driven horizontally off of a level platform at the top of a ramp. The velocity of the car just as it leaves the ramp is 4.36 m/s. The angle of the ramp with respect to the horizontal direction, theta, is 30.8 °. How far does the car travel HORIZONTALLY before landing on the ramp?
How long is the car in the air?
What is the magnitude of the car's velocity just before it lands on the ramp?
How about I do part 1 and then U can do the velocity part?
a very simple sketch (maybe U have this already?)
a horizontal surface with an inclined (ramp) line dropping away
from the horizontal surface at 30.8° downward.
show the horizontal velocity of toy car = 4.36 m/s and
then sketch dotted a HORIZONTAL DISPLACEMENT from the top edge of ramp = "x"
"x" represents the horizontal distance car travels before it hits ramp.
finally sketch dotted a VERTICAL line from the end of "x" down to the ramp line.
Label this vertical line = h {represents vertical height fallen by toy car}
immediately see that:
Tan 30.8° = h/x
also
if we let t = time for toy car to fall to ramp then
4.36t = x
and
from kinematics we know that h = 1/2gt² {vertical height fallen}
substitute
Tan 30.8° = h/x = (0.5)(9.81)t²/4.36t = 4.905/4.36(t) = 1.125t
t = tan 30.8/1.125 = 0.530 s ANS-1
------
to find the speed of the toy car just before it lands U'll need
the two components(vertical & horizontal) of speed and then
combine them with the "hypotenuse rule".
{hint: U already have the horizontal component of speed
& with the value of "h" U can compute the vertical component}
Comments