A ball is launched vertically upward from the edge of a cliff. The ball reaches its maximum height 1.6 seconds after launch. Barely missing the edge of the cliff as it falls downward, the ball strikes the ground 6 seconds after being launched. a) What was the ball's initial velocity? b) What is the maximum height the ball reached above the cliff? c) How tall is the cliff?

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abidemiokin abidemiokin · Answer 1 · 2020-02-04T07:49:48+01:00

Answer: a) 60m/s b) 360m c) 180m

Explanation:

Projectile motion occurs when an object launched into space falls freely under the influence of gravity.

If the time of flight (T) = 2Usin(theta)/g where;

U is the velocity of the object

theta is the angle at which the object was thrown

g is the acceleration due to gravity

Given T = 6 seconds

theta= 90°(object launched upwards)

g = 10m/s²

U= ?

Substituting in the formula we have;

6 = Usin90°/10

60 = Usin90°

since sin90° = 1

U = 60m/s

The balls initial velocity is 60m/s.

b) Maximum Height reached (H) = U²sin²theta/g

Given U = 60m/s, theta = 90° g = 10m/s²

H = 60²(sin90°)²/10

H = 3600/10

H = 360m.

The maximum height reached is 360m

c) at maximum height, v = 0

Using v² = u² - 2gh

0 = 60² - 2(10)h

-3600 = -20h

h = 3600/20

h = 180m

The cliff is 180m tall