Answer:
a. 0.6 m b. 0.385 m c. 3.6 m/s at 287.78° to the horizontal
Explanation:
a. Using s = ut - 1/2gt² for motion under gravity where s = vertical distance = height of table, u = initial vertical velocity of book = 0 m/s, t = time of flight = 0.350 s and g = acceleration due to gravity = 9.8 m/s².
Substituting these these values into s and taking the top of the table as position 0 m, we have.
0 - s = 0t - 1/2gt²
-s = -1/2gt²
s = 1/2gt²
s = 1/2 × 9.8 m/s² × (0.350 s)²
s = 0.6 m
b. Using d = v't where d = horizontal distance from table, v' = horizontal velocity of book = 1.10 m/s and t = time of flight = 0.350 s
d = v't = 1.10 m/s × 0.350 s = 0.385 m
c. Using v² = u² - 2gs where u = initial vertical velocity of book = 0 m/s and g = 9.8 m/s², s = -0.6 m (negative since we are at the bottom and 0 m is at the top)and v = final vertical velocity of book
v² = u² - 2gs
= 0 - 2 × 9.8 m/s² × (-0.6 m)
= 11.76 m²/s²
v = √11.76 m/s
= 3.43 m/s
So, the magnitude of the resultant velocity is V = √(v² + v'²)
= √((3.43 m/s)² + (1.10 m/s)'²)
= √(11.76 m²/s² + 1.21 m²/s²)
= √12.97 m²/s²
= 3.6 m/s
Its direction Ф = tan⁻¹(-v/v') since v is in the negative y direction
= tan⁻¹(-3.43 m/s/1.10 m/s)
= tan⁻¹(-3.1182)
= -72.22°
Ф = -72.22°+ 360 = 287.78° since it is in the third quadrant
