Question

Suppose you throw a ball vertically upward from the flat roof of a tall building. The ball leaves your hand at a point even with the roof railing, with an upward velocity of 15.0 m/s. On its way back down, it just misses the railing. Find (a) the position and velocity of the ball 1.00 s and 4.00 s after it leaves your hand; (b) the velocity of the ball when it is 5.00 m above the railing; and (c) the maximum height reached and the time at which it is reached. Ignore the effects of the air.

Answer 1

(a)

vf_1s = 5.19 m/s

h_1s = 10.095 m

vf_4s = 24.23 m/s

h_4s = 4.49 m (below railing)

(b)

vi = 9.9 m/s

(c)

t = 1.53 s

h = 34.41 m

Step-by-step explanation:

(c)

First, we will use the 1st equation of motion to find the time to attain the highest point:

`v_(f) = v_(i) + gt\\t = (v_(i) - v_(f))/(g)\\`

where,

t = time to attain maximum height = ?

vf = final velocity = 0 m/s (ball momentarily stops at highest point

vi = initial velocity = 15 m/s

g = - 9.81 m/s (for upward motion)

`t = (0\ m/s - 15\ m/s)/(- 9.81\ m/s^2)\\`

t = 1.53 s

Now, for the height attained we will use the 2nd equation of motion:

`h = v_(i)t + (1)/(2)gt^2\\\\h = (15\ m/s)(1.53\ s) + (1)/(2)(9.81\ m/s^2)(1.53)^2\\\\h = 22.93\ m + 11.48\ m`

h = 34.41 m

(b)

using the 3rd equation of motion for a height of 5 m:

`2gh = v_(f)^2 - v_(i)^2\\2(-9.81\ m/s^2)(5\ m) = (0\ m/s)^2 - v_(i)^2 \\v_(i) = √(98.1\ m^2/ s^2)\\`

vi = 9.9 m/s

(c)

At t = 1 s:

`v_(f1s) = v_(i) + gt\\v_(f1s) = 15\ m/s + (-9.81\ m/s^2)(1\ s)\\`

vf_1s = 5.19 m/s

`h_(1s) = v_(i)t + (1)/(2) gt^2\\\\h_(1s) = (15\ m/s)(1\ s) + (1)/(2)(-9.81\ m/s^2){(1\ s)^2}`

h_1s = 10.095 m

At t = 4 s:

Since the ball covers the maximum height of 34.41 m in 1.53 s and then starts moving downward.

Therefore for the remaining 4 s - 1.53 s = 2.47 s, the initial velocity will be 0 m/s at the highest point and the value of g will be positive due to downward motion.

`v_(f4s) = v_(i) + gt\\v_(f4s) = 0\ m/s + (9.81\ m/s^2)(2.47\ s)\\`

vf_4s = 24.23 m/s

`h_(down) = v_(i)t + (1)/(2) gt^2\\\\h_(down) = (0\ m/s)(2.47\ s) + (1)/(2)(9.81\ m/s^2){(2.47\ s)^2}\\\\h_(down) = 29.92\ m`

now, for the position with respect to railing:

`h_(4s) = h - h_(down)\\h_(4s) = 34.41\ m - 29.92\ m\\`

h_4s = 4.49 m (below railing)