1. Scott and jane attempt to jump a gap of 135 meters on scott's Harley. If they leave the jump ramp at a speed of 35m/s and an angle of 45 degrees, do they make the jump?
Please indicate which equation you use, what variable corresponds to each number, and how you got the number that goes with the variable.
Thanks!
Please indicate which equation you use, what variable corresponds to each number, and how you got the number that goes with the variable.
Thanks!
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First, lets look at what we are given and what is required:
Horizontally:
V1x= 35m/s (cos45)
= 24.75m/s
V2x= Unknown
ax= 0m/s2 (since gravity doesn't pull objects horizontally)
dx= Unknown
tx= Unknown
Vertically:
V1y = 35m/s (sin45)
= 24.75m/s
V2y= Unknown
ay= -9.8m/s2 (negative because direction of pull is downwards)
dy= -135 m (negative again)
ty= Unknown
Now, remember that tx = ty, so both times are equal.
The formula we are using is d=(v1)(t) + (0.5)(a)(t)
First, we solve for t in the y-direction (vertically)
dy=(V1y)(t) + (0.5)(ay)(t)2 [squared, not multiplied by two)
-135=(24.75)(t) = (0.5)(-9.8)(t)2
Now, since we cannot go further with this equation, we must rearrange it into quadratic form and use the quadratic equation.
ax2 + bx + c = 0
0= (-4.9)t2 + 24.75t + 135
t = -b +- (squareroot) b2 - 4ac / 2a
(note that it is hard to indicate the square roots and squares, so i trust you are familiar with the quadratic equation, if not look it up and insert the variables)
you should get 2 answers since its the quadratic equation:
-3.3 s OR 8.35 s
HOWEVER, we kow that time cannot be negative so your correct answer is 8.35 seconds.
Now, since we have the time for ty and tx=ty, we can solve for the distance that the motercycle makes:
dx= (V1x)(t) + (0.5)(ax)(t)2 [remember, squared]
dx= (24.75)(8.35) + (0.5)(0)(8.35)2 (now, since there is a zero, the second part cancels out and we get:
dx= (24.75)(8.35)
dx= 206.67m
SO therefore, scott and jane WOULD make the jump on their harley.
These projectile motion questions are a pain ;)
Just practice lots and you'll get the hang of it.
Good luck!
Horizontally:
V1x= 35m/s (cos45)
= 24.75m/s
V2x= Unknown
ax= 0m/s2 (since gravity doesn't pull objects horizontally)
dx= Unknown
tx= Unknown
Vertically:
V1y = 35m/s (sin45)
= 24.75m/s
V2y= Unknown
ay= -9.8m/s2 (negative because direction of pull is downwards)
dy= -135 m (negative again)
ty= Unknown
Now, remember that tx = ty, so both times are equal.
The formula we are using is d=(v1)(t) + (0.5)(a)(t)
First, we solve for t in the y-direction (vertically)
dy=(V1y)(t) + (0.5)(ay)(t)2 [squared, not multiplied by two)
-135=(24.75)(t) = (0.5)(-9.8)(t)2
Now, since we cannot go further with this equation, we must rearrange it into quadratic form and use the quadratic equation.
ax2 + bx + c = 0
0= (-4.9)t2 + 24.75t + 135
t = -b +- (squareroot) b2 - 4ac / 2a
(note that it is hard to indicate the square roots and squares, so i trust you are familiar with the quadratic equation, if not look it up and insert the variables)
you should get 2 answers since its the quadratic equation:
-3.3 s OR 8.35 s
HOWEVER, we kow that time cannot be negative so your correct answer is 8.35 seconds.
Now, since we have the time for ty and tx=ty, we can solve for the distance that the motercycle makes:
dx= (V1x)(t) + (0.5)(ax)(t)2 [remember, squared]
dx= (24.75)(8.35) + (0.5)(0)(8.35)2 (now, since there is a zero, the second part cancels out and we get:
dx= (24.75)(8.35)
dx= 206.67m
SO therefore, scott and jane WOULD make the jump on their harley.
These projectile motion questions are a pain ;)
Just practice lots and you'll get the hang of it.
Good luck!