A worker drops a wrench from the top of a tower 110.5 m tall. What is the veloci

Question

A worker drops a wrench from the top of a tower 110.5 m tall. What is the velocity when the wrench strikes the ground?

A small first-aid kit is dropped by a rock climber who is descending steadily at 1.4 m/s.
(a) After 2.1 s, what is the velocity of the first-aid kit?
(b) After the 2.1 s, how far is the kit below the climber?

A helicopter is ascending vertically with a speed of 5.00 m/s. At a height of 190 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground? [Hint: The package's initial speed equals the helicopter's.]

A small fish is dropped by a pelican that is rising steadily at 0.26 m/s.
(a) After 3.4 s, what is the velocity of the fish?
(b) How far below the pelican is the fish after 3.4 s?

A pitcher throws a baseball horizontally from the mound to home plate. The ball falls 0.997 m (3.27 ft) by the time it reaches home plate 18.3 m (60 ft) away. How fast was the pitcher's pitch?

Explanation / Answer

1) h=110.5m

velocity with which it strikes ground is v

v=sqrt(2*g*h)

v=sqrt(2165.8)

v= 46.538 m/sec

2) initially when he dropped it experiences a pseudo force and goes up with a speed of 1.4 m/sec and then comes down.

time taken to go ho highest point(t1)= u/g

t1=1.4/9.8=0.14285 sec

time it travels downwards(t2) = 2.1- 0.14285= 1.95715 sec

velocity after t2 sec is v

v=gt= 9.8*1.95715= 19.18 m/sec

height travelled from highest point (h1)= 0.5gt^2

h1=0.5*9.8*1.95715^2= 18.769 mts

h2 is the height from the point where he dropped it

h2= u^2/2g

h2= 1.4^2/2*9.8= 0.1 mts

so distance of kit from climber is h1-h2= 18.769-0.1=18.669 mts

**here we considered that the dropper remained at the same position after dropping the kit**

if we consider the dropper to be in the motion then the distance varies, in that case

d= distance travelled by him after dropping the kit

d=2.1*1.4=2.94 mts

distance of kit from climber is 18.669-2.94=15.729 mts

3) h=190m, u=5m/sec

t1= time taken by it to go to the highest point

t1=u/g=5/9.8=0.51 sec

height reached(h1)= u^2/2g

h1= 5*5/2*9.8= 1.2755 mts

new H= h+h1=1.2755+190=191.2755 mts

t=sqrt(2*H/g)

t=sqrt(39.0358)=6.24786 sec

time taken to reach ground is 0.51+6.24786= 6.75786 sec

4) velocity of fish after 3.4 sec is v=u+at

v=0.26+9.8*3.4=33.58 m/sec

d=v^2-u^2/2a= 57.528m

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