Question

A ball is projected upward at time t = 0.0 s from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 80.5 m/s. The velocity of the ball when it is 89 m above the ground is about _____:
Im making this short and simple .

Answer 1

UPWARD JOURNEY
(g is -ive)
u= 80.5m/s
Height reached= u^2/2g= 324m

Height of the ball at topmost point of journey= 414m

DOWNWARD JOURNEY
(g is +ive)
u=0
We need to find the velocity at 89m from ground, which is equivalent to finding velocity at 325 m from the topmost point in downward direction.

v= √2gh= √2*10*325= 80.6= 81 m/sec(Answer)

Answer 2

Answer:

Option D.

Step-by-step explanation:

A ball is projected upward with an initial velocity = 80.5 m per second.

Height of the roof from the ground is 90 m.

Equation that represents the relation between velocity and height when thrown upwards is

$v^{2}=u^{2}-2gh$

Here final velocity v = 0, initial velocity u = 80.5 meter per second and g = 9.8 m per second²

0 = (80.5)² - 2(9.8)h

19.6h = 6480.25

h = $\frac{6480.5}{19.6}=330.625$ meters

Total height of the ball from the ground = 90 + 330.625 = 420.625 m

Now we have to calculate the velocity of the ball at the height = 89 m

Distance covered by the ball falling freely form the point above 420.625

Distance covered by the ball above 89 m = 420.625 - 89 = 331.625 m

We plug in the values in the formula

v² = u² + 2gh

v² = 0 + 2(9.8)(331.625)

v = √6499.85 = 80.62 m

  ≈ 81 m

And the sign notation will be negative

Therefore option D. (-81 m) is the answer.