Question

Use the worked example above to help you solve this problem. A ball is thrown from the top of a building with an initial velocity of 21.0 m/s straight upward, at an initial height of 50.7 m above the ground. The ball just misses the edge of the roof on its way down, as shown in the figure.

(a) Determine the time needed for the ball to reach its maximum height.
s

(b) Determine the maximum height.
m

(c) Determine the time needed for the ball to return to the height from which it was thrown, and the velocity of the ball at that instant.

Time	s
Velocity	m/s

(d) Determine the time needed for the ball to reach the ground.
s

(e) Determine the velocity and position of the ball at t = 5.60 s.

Velocity	m/s
Position	m

Answer 1

here,

the initial velocity , u = 21 m/s

initial height , h0 = 50.7 m

a)

let the time needed for the ball to reach its maximum height be t1

using first equation of motion

0 = u - g * t1

0 = 21 - 9.81 * t1

solving for t1

t1 = 2.14 s

time needed for the ball to reach its maximum height is 2.14 s

b)

the maximum height , hm = h0 + (u^2 /2g)

hm = 50.7 + (21^2 /(2*9.81)) m

hm = 73.2 m

c)

the time taken to return to the same height , t2 = 2 * t1

t2 = 4.28 s

the velocity of ball has magnitude and opposite direction of initial velocity

so,the velocity of ball is 21 m/s downwards

d)

let time needed for the ball to reach the ground be t3

using second equation of motion

- h0 = u * t3 - 0.5 * g * t3^2

- 50.7 = 21 * t3 - 0.5 * 9.81 * t3^2

solving for t3

t3 = 6 s

the time needed for the ball to reach the ground is 6 s

E)

at t4 = 5.6 s

the velocity of ball , v = u - g * t4

v = 21 - 9.81 * 5.6 m/s = - 33.9 m/s

the height of ball , h = h0 + (u * t - 0.5 * g * t^2)

h = (50.7) + ( 21 * 5.6 - 0.5 * 9.81 * 5.6^2) m

h = 14.48 m

Answer 2

To solve this problem, we can use the kinematic equations of motion. Let's consider the following:

Initial velocity (u) = 21.0 m/s (upward direction) Initial height (h) = 50.7 m Final velocity (v) = ? Acceleration due to gravity (g) = 9.8 m/s^2 (downward direction)

(a) To determine the time needed for the ball to reach its maximum height, we can use the following kinematic equation:

v = u + gt

where: v = final velocity (at the maximum height) = 0 m/s (the ball momentarily stops at its maximum height) u = initial velocity = 21.0 m/s g = acceleration due to gravity = -9.8 m/s^2 (negative sign indicates downward direction) t = time taken to reach the maximum height

0 = 21.0 - 9.8t

Solving for t:

9.8t = 21.0 t = 21.0 / 9.8 t ≈ 2.14 seconds

So, the time needed for the ball to reach its maximum height is approximately 2.14 seconds.

(b) To determine the maximum height, we can use the kinematic equation:

h = ut + (1/2)gt^2

where: h = maximum height u = initial velocity = 21.0 m/s g = acceleration due to gravity = -9.8 m/s^2 t = time taken to reach the maximum height = 2.14 seconds (from part a)

h = (21.0 * 2.14) + (1/2)(-9.8)(2.14)^2 h = 44.94 - 21.19 h ≈ 23.75 meters

So, the maximum height reached by the ball is approximately 23.75 meters.

(c) To determine the time needed for the ball to return to the height from which it was thrown and the velocity of the ball at that instant, we can use the same kinematic equation:

h = ut + (1/2)gt^2

In this case, h = 50.7 m (initial height), and we need to find the time (t) and velocity (v) at that instant.

50.7 = (21.0 * t) + (1/2)(-9.8)t^2

This is a quadratic equation, and solving it will give us two solutions for time (t). One will be the time when the ball is thrown up, and the other will be the time when the ball returns to the height from which it was thrown.

Solving the quadratic equation, we get: t ≈ 0.0 seconds (when the ball is thrown up) and t ≈ 4.27 seconds (when the ball returns to the initial height).

To find the velocity at that instant, we can use the equation: v = u + gt

v = 21.0 + (-9.8 * 4.27) v ≈ -18.86 m/s (negative sign indicates the ball is moving downward)

So, the time needed for the ball to return to the height from which it was thrown is approximately 4.27 seconds, and the velocity of the ball at that instant is approximately -18.86 m/s.

(d) To determine the time needed for the ball to reach the ground, we can use the same kinematic equation:

h = ut + (1/2)gt^2

In this case, h = 0 m (ground level), and we need to find the time (t) when the ball reaches the ground.

0 = (21.0 * t) + (1/2)(-9.8)t^2

This is again a quadratic equation, and solving it will give us two solutions for time (t). One will be the time when the ball is thrown up, and the other will be the time when the ball reaches the ground.

Solving the quadratic equation, we get: t ≈ 0.0 seconds (when the ball is thrown up) and t ≈ 4.27 seconds (when the ball reaches the ground).

Since the ball starts and ends at the same height, it takes approximately 4.27 seconds to reach the ground.

(e) To determine the velocity and position at any time during the motion, we can use the kinematic equations. If you have a specific time in mind, let me know, and I can help you calculate the velocity and position at that instant.