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

b) returns freshwater to the earth.

c) returns freshwater to the earth.c) occurs after condensation.

d) returns freshwater to the earth.c) occurs after condensation.d) roughly equal to the amount of evaporation.

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ammonium chloride & tallow