an object is thrown vertically upwards with an initial velocity of 60 metre per second. calculate height attained by an object , time taken to reach the topmost point ,the velocity of the object when it is halfway up.
Answers
Given :
An object is thrown vertically upwards with an initial velocity of 60 metre per second
initial velocity= u= 60 m/s
g=10 m/s²
To find :
Height attained by an object.=h=?
time taken to reach the topmost point = time of ascent=ta=?
Velocity of object when it is halfway.
Solution :
Lets find out time taken to reach the maximum height = time of ascent :
Formula : t= u/g [ From first equation of motion for vertically projected object]
t= 60/10
= 6 sec
∴ Time taken to reach maximum height is 6 sec
Lets find out height attained by object:
h(max)= u²/ 2g [from second equation of motion for vertically projected object]
= 60x60/2x10
=180 m
∴ The maximum height attained by an object is 180m.
Now lets find out velocity of an object at its halfway [ 90m]
from third equation of motion,
V² - U² = 2gs
v² = 60 x 60 +2 x10x 90
=3600 +1800
=5400
v=√5400= 73.48 m/s
Answer:
Given :
An object is thrown vertically upwards with an initial velocity of 60 metre per second
initial velocity= u= 60 m/s
g=10 m/s²
To find :
Height attained by an object.=h=?
time taken to reach the topmost point = time of ascent=ta=?
Velocity of object when it is halfway.
Solution :
Lets find out time taken to reach the maximum height = time of ascent :
Formula : t= u/g [ From first equation of motion for vertically projected object]
t= 60/10
= 6 sec
∴ Time taken to reach maximum height is 6 sec
Lets find out height attained by object:
h(max)= u²/ 2g [from second equation of motion for vertically projected object]
= 60x60/2x10
=180 m
∴ The maximum height attained by an object is 180m.
Now lets find out velocity of an object at its halfway [ 90m]
from third equation of motion,
V² - U² = 2gs
v² = 60 x 60 +2 x10x 90
=3600 +1800
=5400
v=√5400= 73.48 m/s