Abaseball is hit with a speed of 28.0  m/s  at an angle of 42.0  ∘  . it lands on the flat roof of a 12.0  m  -tall nearby building. if the ball was hit when it was 1.5  m  above the ground, what horizontal distance does it travel before it lands on the building?  

consider the motion of the ball along the vertical direction or Y-direction.

Y₀ = initial position of the ball at the time of launch = 1.5 m

Y = final position of the ball at the time of landing  = 12 m

v₀ = initial velocity at the time of launch = 28 Sin42 = 18.74 m/s

a = acceleration = - 9.8 m/s²

t = time taken

using the kinematics equation

Y = Y₀ + v₀ t + (0.5) a t²

12 = 1.5 + (18.74) t + (0.5) (- 9.8) t²

t = 3.14 sec

consider the motion of the ball along the horizontal direction or X-direction.

X₀ = initial position of the ball at the time of launch = 0 m

X = final position of the ball at the time of landing  = ?

v₀ = initial velocity at the time of launch = 28 Cos42 = 20.8 m/s

a = acceleration =0 m/s²

t = time taken = 3.14 sec

using the kinematics equation

X = X₀ + v₀ t + (0.5) a t²

X = 0 + (20.8) (3.14) + (0.5) (0) (3.14)²

X = 65.3 m

hence the ball travels 65.3 m before it lands on building

are their multiple choices ?

explanation: