Question

Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a speed of 5.30 m/s from a height h = 1.20 m. Marble 2 is launched from ground level with a speed of 7.50 m/s at an angle = 45.0 °
above the horizontal. Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates

Answer 1

Explanation:

Horizontal component

cos(45^{0})=\frac{V_{2o}}{V_{2ox}} \\ V_{2ox}=V_{2o}cos(45^{0})\\ V_{2ox}=5.94\frac{m}{s}cos(45^{0}) \\V_{2ox}=4.20\frac{m}{s}cos(450)=V2oxV2oV2ox=V2ocos(450)V2ox=5.94smcos(450)V2ox=4.20sm

Vertical component.

sin(45^{0})=\frac{V_{2o}}{V_{2oy}} \\ V_{2oy}=V_{2o}sin(45^{0})\\ V_{2oy}=5.94\frac{m}{s}sin(45^{0}) \\V_{2oy}=4.20\frac{m}{s}sin(450)=V2oyV2oV2oy=V2osin(450)V2oy=5.94smsin(450)V2oy=4.20sm

When there is no gravity.

Marble 1 will follow a horizontal path.

Marble 2 moves from the ground to the height of marble 1 in parallel, therefore it is necessary to calculate the time it takes to travel the height h =0.950m

The time it takes for marble 2 to collide with marble 1 is given by:

v_{oy}=\frac{Y}{t} \\t=\frac{Y}{V_{oy}}voy=tYt=VoyY

Where:

Vertical velocity component V_{oy}=4.20\frac{m}{s}Voy=4.20sm

Distance Y=h--->h=0.950mY=h−−−>h=0.950m

time tt

Numerically evaluating

t=\frac{0.950m}{4.20\frac{m}{s}}\\t=0.226st=4.20sm0.950mt=0.226s

The X coordinate where they collide is given by:

X=X_{o}+V_{1ox}*tX=Xo+V1ox∗t

Where

Marble Horizontal Velocity 1 V_{1ox}=4.20\frac{m}{s}V1ox=4.20sm

Initial position X_{o}=0mXo=0m

time t=0.226st=0.226s

Evaluating numerically. X=0m+4.20\frac{m}{s}*0.226s=0.949mX=0m+4.20sm∗0.226s=0.949m

Finally both marbles collide at a height of h = 0.950m and at a horizontal distance of x =0.949m

The coordinate (x, y) where they collide is equal to \boxed{(0.949,0.950)m}(0.949,0.950)m