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nordsb [41]
3 years ago
13

An inflatable raft (unoccupied) floats down a river at an approximately constant speed of 5.6 m/s. A child on a bridge, 71 m abo

ve the river, sees the raft in the river below and attempts to drop a small stone onto the raft. The child releases the stone from rest. In order for the stone to hit the raft, what must be the horizontal distance between the raft and the bridge when the child releases the stone?
Physics
1 answer:
saveliy_v [14]3 years ago
3 0

Answer:

21.28 m

Explanation:

height, h = 71 m

velocity of raft, v = 5.6 m/s

let the time taken by the stone to reach to raft is t.

use second equation of motion for stone

h = ut + \frac{1}{2}at^{2}

u = 0 m/s, h = 71 m, g = 9.8 m/s^2

71 = 0 + 0.5 x 9.8 x t^2

t = 3.8 s

Horizontal distance traveled by the raft in time t

d = v x t = 5.6 x 3.8 = 21.28 m

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3 years ago
A hockey player strikes a puck, giving it an initial velocity of 14.0 m/s in the positive x-direction. The puck slows uniformly
Advocard [28]

a) -1.71 m/s^2

b) 7.7 m/s

c) 4.39 s

Explanation:

a)

The acceleration of an object is the rate of change of velocity of an object.

In this problem, the acceleration of the puck can be found using the following suvat equation:

v^2-u^2=2as

where:

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the puck in this problem:

u = 14.0 m/s

v = 6.50 m/s

s = 45.0 m

So, the acceleration is:

a=\frac{v^2-u^2}{2s}=\frac{6.50^2-14.0^2}{2(45.0)}=-1.71 m/s^2

b)

The velocity of the puck at time t can be found by using another suvat equation:

v=u+at

where

u is the initial velocity

a is the acceleration

t is the time elapsed

v is the final velocity

Here, we have:

u = 14.0 m/s

a=-1.71 m/s^2 (found in part a)

Therefore, the velocity of the puck after t = 3.70 s is:

v=14.0+(-1.71)(3.70)=7.7 m/s

c)

Here we want to find the time taken for the puck to travel a distance of

s = 45.0 m

To solve this part, we can use again the suvat equation:

v=u+at

Where in this case, we use:

u = 14.0 m/s is the initial velocity

v = 6.50 m/s is the final velocity when the puck has travelled 45.0 m (this information is given in the question)

a=-1.71 m/s^2 is the acceleration (found in part a)

Therefore, by re-arranging the equation, we find the time taken to cover 45.0 m:

t=\frac{v-u}{a}=\frac{6.50-14.0}{-1.71}=4.39 s

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Answer:

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