Answer:
a. 6.39 s b. 255.6 m c. 62.6 m/s
Explanation:
a. Find the time it takes to hit the ground.
Using y = ut + 1/2gt² where y = height of cliff = 200 m, u = initial vertical velocity of ball = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for ball to hit the ground.
So, substituting the values of the variables into the equation, we have
200 m = 0 m/s × t + 1/2 × 9.8 m/s² × t²
200 m = 0 + (4.9 m/s²)t²
200 m = (4.9 m/s²)t²
dividing both sides by 4.9 m/s², we have
t² = 200 m ÷ 4.9 m/s²
t² = 40.82 s²
taking square root of both sides, we have
t = √(40.82 s²)
t = 6.39 s
b. Find the RANGE, distance traveled horizontally.
The range is the horizontal distance traveled by the ball, R = vt where v = horizontal velocity of ball = 40 m/s and t = time taken for ball to hit the ground = 6.39 s
So, R = vt
= 40 m/s × 6.39 s
= 255.6 m
c. Find the final vertical velocity.
Using v = u - gt where u = initial vertical velocity of ball = 0 m/s, v = final velocity of ball, g = g = acceleration due to gravity = 9.8 m/s² and t = time taken for ball to hit the ground = 6.39 s.
So, substituting the values into the equation, we have
v = 0 m/s - 9.8 m/s² × 6.39 s
v = -62.62 m/s
v ≅ -62.6 m/s
