The ball as launched horizontally off a 200m cliff. The initial velocity is 40m/s. Find the time it takes to hit the ground.
Find the RANGE, distance traveled horizontally.
Find the final vertical velocity.

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

the answer is b

explanation:

the answer is 12 as i may remember

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