# A stone is thrown upward from ground level. The initial speed is 32 ft/sec. (a) In how many seconds will the stone hit the ground?(b) How high will it go? (c) With what minimum speed should the...

A stone is thrown upward from ground level. The initial speed is 32 ft/sec. (a) In how many seconds will the stone hit the ground?

(b) How high will it go? (c) With what minimum speed should the stone be thrown so as to reach a height of at least 36 ft?

### 2 Answers | Add Yours

The stone thrown upwards will slow down under the influence of gravity till it reaches 0 velocity/speed and will again start falling down with increasing velocity in downward direction till it hits the ground. The magnitude of velocity at the time of hitting the ground will be same as the velocity at which the stone was thrown. The acceleration of the stone due to gravity is equal to 32 feet/s^2.

Thus we can find the answer to the given questions as follows:

(a)

Acceleration due to gravity = a = 32 feet/second^2

Upward velocity = u = 32 feet/s

Final velocity when the stone reaches its highest point = v = 0

Time take t o reach highest point = t = (u -v)/a = 32/32 = 1 s

The stone will take the same time to fall down and hit the ground. Therefore:

The taken by stone to hit the ground = 1 + 1 = 2 seconds

(b)

Distance travelled by stone on upward travel = d = ut/2

= 32*1/2 = 16 feet

The stone will go up 16 feet high.

(c)

Distance of highest point reached by stone is given by:

Distance travelled = (a*t^2)/2

When the stone to travel 36 feet high:

36 = (32*t^2)/2

t^2 = 36/16 = 2.25

Therefore:

t = 1.5 s

As the stone takes 1.5 seconds to reach 0 velocity:

Initial velocity = t*a = 1.5*32 = 48 feet/s

Answer:

Minimum required speed for stone to reach a height of at least 36 feet = 48 feet/s

The initial velocity u of the stone is 32.

The stone reaches a final velocity v = 0 at a time t given by v= u=gt, where g is the acceleration due to gravity. So 0 = 32-gt. Or t = 32/g. We assume g = 32 ft/s^2.

Therefore t = 32/g = 32/32 = 1second.

We answer (b) first:

b)

The equation of motion for the height of the stone is h = ut-(1/2)gt^2, where u = intial velocity = 32ft/s. We Put t = 1 in the equation h = ut - (1/2)gt^2, when the final velocity is zero at the highest point and the stone begins to fall:

h = 32*1-(1/2)32(1^2) = 32 - (1/2)32*1^2 = 32-16 = 16 ft.

So the stone reaches a height of 16 ft .

a)

The stone takes a time of 1 secod to reach maximum height of 16 ft. Then it falls. The falling time is also the same as the time to reach the highest point. So the stone takes 2 seconds to reach the ground from the time it was thrown up.

c)

To find the initial speed of the stone so that it reaches a height of at least 36 ft. So h > 36.

We use the equation of motion t v^2-u^2 = 2gh where u is the initial velocity and v = final velocity = 0 at the highest point and g is the acceleration due to gravity and h = height and g = -32 as it is in opposite direction to the velocity.

In this case h > 36.

Therefore 2*(-g) 36 > 0^2- u^2. g= 32.

2*32*36 < u^2.

u = sqrt(2*32*36) = 8*6 = 48.ft/s.

So the miminmum velocity of the stone to reach at least 36 ft high is 48ft/s.

So 2gs > v^2 -u^2