A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground?NCERT Class IXSciences - Main Course BookChapter 8. Motion
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When a body having uniformly accelerated motion travels in a straight line the relationship between its initial velocity(u) final velocity (v) , acceleration(a) during its motion , distance covered (s) by it in a certain time interval( t ) are called equations of motion.
There are three equations for the motion of a body moving with a uniform acceleration.
These three equations are:
v=u+at
S= ut+ ½ at²
v²=u² +2as.
To solve numerical problem on motion we should remember the following:
⇒If a body starts from the state of rest then u= 0⇒
If a body stops or comes to the state of rest then v= 0
⇒If a body moves with a uniform (constant) velocity then a=0.
=========================================================
Solution:
Given:
Initial velocity,u=0 m/s [since the ball is just at rest when dropped]
Distance,(height) (s)=20 m
Acceleration,a= 10 m /s²
Final velocity,v=?
Time taken to strike the ground(t)= ?
(a)For Final velocity(v) v²=u²+2as [3rd eq. of motion]
⇒v²=0+2×10 ×20
⇒v²= 20×20
⇒v²=400
⇒v=√400
⇒v=20 m/s
(b) For time (t) v=u+at [1st eq. of motion]
⇒20 =0+10(t)
⇒20 = 10t
t= 20/10
t= 2 sec
Therefore, the final velocity of the Ball when it strikes the ground is 20 m/s
The ball will strike the ground after 2 sec
=========================================================
Hope this will help you...
There are three equations for the motion of a body moving with a uniform acceleration.
These three equations are:
v=u+at
S= ut+ ½ at²
v²=u² +2as.
To solve numerical problem on motion we should remember the following:
⇒If a body starts from the state of rest then u= 0⇒
If a body stops or comes to the state of rest then v= 0
⇒If a body moves with a uniform (constant) velocity then a=0.
=========================================================
Solution:
Given:
Initial velocity,u=0 m/s [since the ball is just at rest when dropped]
Distance,(height) (s)=20 m
Acceleration,a= 10 m /s²
Final velocity,v=?
Time taken to strike the ground(t)= ?
(a)For Final velocity(v) v²=u²+2as [3rd eq. of motion]
⇒v²=0+2×10 ×20
⇒v²= 20×20
⇒v²=400
⇒v=√400
⇒v=20 m/s
(b) For time (t) v=u+at [1st eq. of motion]
⇒20 =0+10(t)
⇒20 = 10t
t= 20/10
t= 2 sec
Therefore, the final velocity of the Ball when it strikes the ground is 20 m/s
The ball will strike the ground after 2 sec
=========================================================
Hope this will help you...
Answered by
14
ɢɪᴠᴇɴ
ɪɴɪᴛɪᴀʟ ᴠᴇʟᴏᴄɪᴛʏ(ᴜ) = 0 ᴍ/ꜱ
ᴀᴄᴄᴇʟᴇʀᴀᴛɪᴏɴ (ᴀ) = 10 ᴍ/ꜱ²
ᴛɪᴍᴇ(ꜱ) = 20 ᴍ
ᴡᴇ ᴋɴᴏᴡ
ᴠ² = ᴜ² + 2ᴀꜱ
ᴘᴜᴛᴛɪɴɢ ᴠᴀʟᴜᴇꜱ:
ᴠ² = 0² + 2 (10 x 20)
ᴠ² = 400
ᴠ = 20 ᴍ/ꜱ
ꜰᴏʀ ᴛɪᴍᴇ:
ᴠ = ᴜ + ᴀᴛ
ᴛ = ᴠ - ᴜ/ᴀ
ᴛ = (20-0)10
= 20/10
= 2
∴ ꜱᴛʀɪᴋɪɴɢ ᴠᴇʟᴏᴄɪᴛʏ = 20 ᴍ/ꜱ
∴ ꜱᴛʀɪᴋɪɴɢ ᴛɪᴍᴇ = 2 ꜱᴇᴄᴏɴᴅꜱ
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