Answer:
a. 50 m/s b. 0 m/s c. 50 m/s d. 4.04 s e. -39.6 m/s f. 63.78 m/s g. 202 m
Explanation:
a. What is the initial horizontal velocity?
Since the object is launched horizontally, it initial horizontal velocity is 50 m/s
b. What is the initial vertical velocity?
Since the object is launched horizontally, it has no initial vertical component. So, its initial vertical velocity is 0 m/s
c. What is the final horizontal velocity?
Its final horizontal velocity is 50 m/s since no force acts on it in the horizontal direction to change its value.
d. How much time did it take for the object to hit the ground?
We use the equation s = ut - 1/gt² since the object is falling under gravity where u = initial vertical velocity = 0 m/s, s = height of cliff = 80 m, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the object to hit the ground.
s = ut - 1/2gt²
80 m = 0 × t - 1/2 × -9.8 m/s² × t²
80 m = 4.9 m/s² × t²
t² = 80 m ÷ 4.9 m/s²
t² = 16.33 s²
t = √(16.33 s²)
t = 4.04 s
e. What is the final vertical velocity?
Using v = u + at where u = initial vertical velocity = 0 m/s, v = final vertical velocity, a = acceleration = -g = -9.8 m/s²and t = time it takes object to reach the ground = 4.04 s.
Substituting these values into the equation, we have
v = u + at
v = 0 m/s + (-9.8 m/s²) × 4.04 s
v = -39.6 m/s
f. What is the final resultant speed?
The final resultant speed v' is the resultant of the final horizontal velocity and the final vertical velocity. Let u' = final vertical velocity = 50 m/s.
v' = √(u'² + v²)
v' = √((50 m/s)² + (-39.6 m/s)²)
v' = √(2500 m²/s² + 1568.16 m²/s²)
v' = √(4068.16 m²/s²)
v' = 63.78 m/s
g. How far from the base of the cliff will the projectile land?
The distance from the base of the cliff, d where the projectile lands is
d = u't where u' = horizontal velocity = 50 m/s and t = time it takes object to land = 4.04 s
d = 50 m/s × 4.04 s
d = 202 m
