A marble is shot out of a gun. The following variables are recorded. There is no

Question

A marble is shot out of a gun. The following variables are recorded. There is no exit or final velocity given.
Distance from wall: 647 cm
Launch Angle: 35 Degrees
Height of Impact on Wall: 38.5 cm
Height of Lauch: 6.5 cm
Time to hit wall: 0.76 seconds
Acceleration due to gravity: 9.8 m/s^2
Using the info provided find the exit velocity of the marble from the gun.
BONUS:
Find the distance and angle the gun must be shot at in order to shoot the marble through a hole 2.44m above the ground at its maximum trajectory.

Explanation / Answer

in horizontal.

d_x = vx t

6.47 m = (v0x) (0.76 s)

v0x = 8.51 m/s

In vertical, yf - yi = v0y t + ay t^2 / 2

0.385 m - 0.065 = (v0y)(0.76) + (-9.8 x0.76^2 /2 )

v0y = 4.14 m/s

so v0 = sqrt(v0x^2 + v0y^2) = 9.46 m/s .......Ans

Bonus:

we want max height to be 2.44 m from ground.

so H = 2.44 - 0.065 = 2.375

H = (v0 sin(theta))^2 / 2 g

2.375 = (9.46 sin(theta))^2 / (2 x 9.8)

sin(theta) = 46.2 deg ........Ans

and R = v0^2 sin(2 theta)/g

R = (9.46^2)sin(2*46.3) / 9.8

R = 9.12 m

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