Question

A baseball thrown at an angle of 55.0 ∘ above the horizontal strikes a building 18.0 m away at a point 5.00 m above the point from which it is thrown. Ignore air resistance.

Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown).

Find the magnitude of the velocity of the baseball just before it strikes the building.

Find the direction of the velocity of the baseball just before it strikes the building.

Answer 1

x = Vo*t *cos

t = x/(Vo cos)

y = Vo*t*sin + 1/2*g*t²

y = Vo (x/(Vo*cos)) sin +1/2*g* (x/(Vo*cos))²

y = x *tan + 1/2*g *(x/(Vo*cos))²

y = (18) tan55 + 1/2* (-9.8)*(18/(Vo*cos55))² = 5

25.70 - 4826.96/Vo² = 5

Vo = 15.27 m/s

(b)

t = x/(Vo*cos)

t = 18/(15.27 cos 55)

t = 2.05 sec.

Vx = Vo*cos = 15.27 *cos55

Vx = 8.75 m/s (forward)

Vy = Vo*sin + g*t = 15.27*sin 55 + (-9.8)(2.05)

Vy = -7.59 m/s (downward)

resultant of ball's velocity is:

Vr = sqrt (Vx² + Vy²) = sqrt( (8.75)² + (-7.59)²)

Vr = 11.58 m/s

(c)

to determine direction of ball's velocity when hits the building,

tan = Vy/Vx = -7.59/8.75 = -0.86

= - 40.69^o