Question

3- Your friend drops a tennis ball from the window of her apartment (66.1 m above the ground). One second later you throw a rock (as a projectile) from the ground as shown in the figure. The rock and the ball collide at x = 32.6 m, y = 22.0 m measured from where the rock is thrown as shown in the figure.

a) Determine how many seconds after the tennis ball is released the collision takes place.

b) Determine the velocity of the tennis ball at the time of its collision with the rock.

c) Determine the initial velocity of the rock in its component form.

Answer #1

A) the distance travelled by the ball before making collision
with the rock is 66.1-22 = 44.1 m

then ball is a freely falling body case

initial speed is u = 0 m/s

Distance travelled is s= 44.1 m

accelaration due to gravity is a = g = 9.8 m/s^2

apply s = (u*t)+(0.5*g*t^2)

44.1 = 0+(0.5*9.81*t^2)

t = sqrt(44.1/(0.5*9.81)) = 3 sec

B) after 3sec

v = u+(g*t) = (9.81*3)= 29.43 m/s

C) vx = x/t = 32.6/3 = 10.86 m/s

y = (vy*t)+(0.5*ay*t^2)

22 = (vy*3)-(0.5*9.81*3^2)

vy = 22.04 m/s

in component's form

v = 10.86 i + 22.04 j

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