Question

5. I am standing on a cliff that puts my throwing arm about 25 m above the ground level.

(a) If I can throw a baseball at 12 m/s, what?s the maximum angle I could throw it at so that it would always stay below 30 m above the ground?

(b) Write the initial velocity of the baseball from the previous part in vector format.

c) How far from the base of the cliff is the baseball going to land if I throw it like this.

I know the answers but I am looking for the work. A=56 degress, 24m

Answer 1

here,

h0 = 25 m

initial speed , u = 12 m/s

a)

let the angle be theta

the maximum height , hmax = ( u * sin(theta))^2 /(2 * g)

30 - h0 = ( u * sin(theta))^2 /(2 * g)

30 - 25 = ( 12 * sin(theta))^2 /( 2 * 9.81)

solving for theta

theta = 56 degree

b)

the initial velocity , u = 12 * ( cos(theta) i + sin(theta) j)

u = 6.71 i m/s + 9.95 j m/s

c)

let the time taken to hit the ground be t

h0 = -uy * t + 0.5 * g * t^2

25 = - 9.95 * t + 0.5 * 9.81 * t^2

solving for t

t = 3.48 s

the horizontal distance , x = ux * t

x = 6.71 * 3.48 m = 24 m