# A ball is thrown vertically upward with a speed of 24.0m/s. (a) How high does it rise? (b) How long does it take to reach its highest point? (c) How long does the ball take to hit the ground after...

A ball is thrown vertically upward with a speed of 24.0m/s.

(a) How high does it rise?

(b) How long does it take to reach its highest point?

(c) How long does the ball take to hit the ground after it reaches its highest point?

(d) What is its velocity when it returns to the level from which it started?

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### 2 Answers

Start with finding the time it takes it to reach the highest point at which the velocity = 0 and it will start reversing its path due to gravity.

This can be found from the equation v = u - gt (note gravity works against the ball hence the -ve sign). v is final velocity which is 0, g is acceleration of gravity 9.81 m/s^2 , u is initital velocity of 24 m/s and t is the time it takes to attain a zero velocity

Therefore,

0 = 24 - 9.81 *t

solving, t = 24/9.81 = 2.45 sec

(a) The vertical distance it travels is how high it will go, which is represented by

s = ut - 1/2 *g*t^2 (note the -ve sign for g here as well)

= 24 * 2.45 - 1/2* 9.81 * 2.45^2 = 29.35m

(b) Time t = 2.45 sec al already calculated

(c) This can be obtained from same equation s = ut + 1/2 *g*t^2, except in this case g is +ve as gravity works for the fall and initial verlocty u = 0

knowing the distance traveled is s = 29.35 m (same distance it traveled upwards)

29.35 = 0*t + 1/2 * 9.81 * t ^2

Solving, t = sqrt (29.35/(0.5*9.81)) - 2.45 sec

(d) Final verlocity v = u + gt (note +ve for gravity)

= 0 + 9.81*2.45

= 24 m/s

Which is equal to the velocity it was thrown upwards and serves as a cross check for the answer. In a frictionless environment, no energy is lost and hence the ball will retain same energy as it had when it was thrown upwards, which explains why it has same velocity at return.

**a)**

To calculate the height in a vertical motion, we apply the following formula:

h = (v^2 – v0^2)/2g **(1)**

where:

v → is the velocity at any instant of movement. At the highest point, the speed of the body is zero.

v0 → is the initial velocity.

g → is the acceleration of gravity. Since the movement is up, we take g with a negative sign.

h = (v^2 – v0^2)/2g = (0 – 24^2)/2(-9.8) = (-576)/(-19.6) = 29.39 m

**b)**

The time to reach the maximum height we can calculate, by solving the equation of speed.

v = v0 - gt **(2)**

Solving for t, we have:

t = (v - v0)/g = (0 – 24)/(-9.8) = 2.45 s

**c)**

The time to reach the floor, after reaching the maximum height, is calculated by solving the equation **(2)**. In this case we can find the final velocity using equation **(1)**, applied in part **(a)**. The initial velocity is zero.

h = (v^2 – v0^2)/2g

**v = sqrt (2gh + v0^2) = sqrt (2(9.8)(29.39) + 0) = 24 m/s**

Solving equation (2) for t:

t = (v - v0)/g = (24 – 0)/9.8 = 2.45 s

**d)**

This response is included, in bold, in the part **(c)**. When the ball returns to the level from which it started, the speed is 24 m/s. This is a consequence of the law of conservation of energy.