Projectile?

Question

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

all serious answers are welcomed. thankyou :)

Answer 1

Horizontal component of distance is x=cos(45)(30 m/s)(time)

Vertical component of distance is y=sin(45)(30 m/s)(time) - (1/2)(9.81 m/s^2)(time)^2

First solve for how long it will be in the air. It start 2 meters off the ground so we must solve for time when y = -2 (that's when it will hit the ground):

-2=sin(45)(30 m/s)(t) - (1/2)(9.81 m/s^2)(t)^2
t= about 4.417 sec

Now we find horizontal distance when the guy hits the ground:

x=cos(45)(30 m/s)(4.417 sec)
x= about 93.699 meters

That's about one football field!

Now we must use the pythagorean theorem to add the distances. He want 2 meters down and 93.699 meters out.

sqrt(2^2+93.699^2) = 93.720 meters is the total distance.

Answer 2

Do you mean the horizontal distance? Start by using trigonometry to determine the vertical component of velocity and the horizontal component of velocity. Use y = V_0y*t + 0.5*g*t^2, where V_0y is the initial vertical velocity and gravity is negative (with the initial upwards velocity positive) to determine how much time must pass for the motorcycle to fall from the initial 2m height down to zero. Then, given that time and ignoring air resistance, just calculate x = V_0x*t, where V_0x is the initial (and constant) horizontal velocity.

Answer 3

break into horizonal and vertical velocitiy components.(assume x is horizonal velocity and y is vertical velocity.)
Vy=sin(45)(30)=21.21m/s
Vx=cos(45)(30)=21.21m/s
calculate the velocity when the stuntman reaches the ground.
Vf^2=2ad+Vi^2
Vf^2=2(-9.8)(-2)+(21.21)^2
Vf^2=39.2+449.8641
Vf^2=489.06
Vf= -22.11m/s

calculate the time to reach the ground
Vf=at+Vi
-22.11=(-9.8)t+21.21
-43.32=(-9.8)t
t=4.42s

calculate the horizonal distance
x=vt
x=(21.21)(4.42)
x=93.75m

Answer 4

break into horizonal and vertical velocitiy components.(assume x is horizonal velocity and y is vertical velocity.)
Vy=sin(45)(30)=21.21m/s
Vx=cos(45)(30)=21.21m/s
calculate the velocity when the stuntman reaches the ground.
Vf^2=2ad+Vi^2
Vf^2=2(-9.8)(-2)+(21.21)^2
Vf^2=39.2+449.8641
Vf^2=489.06
Vf= -22.11m/s

calculate the time to reach the ground
Vf=at+Vi
-22.11=(-9.8)t+21.21
-43.32=(-9.8)t
t=4.42s

calculate the horizonal distance
x=vt
x=(21.21)(4.42)
x=93.75m

Answer 5

when a body is projected horizontally other than 90degrees along the path is called projectile

Answer 6

As given above, about 94 mts.

Answer 7

who r solving this

go and save his life