A ball is gently dropped from a height of 20m

Distance ball covers, S = 20 m acceleration, a = 10Kmh -1= 10 m/s initial velocity, u = 0 (at rest) When ball strikes, its final velocity, According to 3rd law of motion v 2= u 2+ 2as v 2= 0 + 2(10)(20) v = 20 m/s. According to 1 st Law of motion v = u + at Time taken by ball to strike the ground, t, 20 = 0 + 10 (t) ∴ t = 2s ∴ 2s is required to ball for striking the ground and velocity is 20 m/s

Solution Here, u = 0 , a = 10 m s − 2 , s = 20 m . 2 a s = v 2 − u 2 ∴ 2 × 10 m s − 2 × 20 m = v 2 − 0 ∴ v 2 = 400 m 2 s − 2 t h e r e o e v = 20 m s − 1 v = u + a t ∴ 20 m s − 1 = 0 + 1 0 m s − 2 × t ∴ 20 m s − 1 = 10 m s − 2 × t ∴ t = 20 m s − 1 10 m s − 2 = 2 s The ball will strike the ground with a velocity of 20 m s − 1 . It will strike the ground after 2s.

Fig shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions: ( a ) Which of the three is travelling the fastest? ( b ) Are all three ever at the same point on the road? ( c ) How far has C travelled when B passes A? ( d ) How far has B travelled by the time it passes C? 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? Updated On Apr 26, 2023 Topic Motion Subject Science Class Class 9 Answer Type Text solution:1 Video solution: 1 Upvotes 93 Avg. Video Duration 6 min

Step 1: Given Data Initial velocity u = 0 Distance traveled by the ball S = 20 m Uniform acceleration a = 10 m s - 2 Let the final velocity be v. Let the time taken be t. Step 2: Calculate the FInal Velocity According to Newton's third law of motion v 2 = u 2 + 2 a S ⇒ v 2 = 0 + 2 × 10 × 20 ⇒ v = 20 m / s Step 3: Calculate the Time Taken According to Newton's first law of motion v = u + a t ⇒ t = v - u a ⇒ t = 20 - 0 10 = 20 m s 10 m s 2 ⇒ t = 2 s Hence, the ball will strike the ground at a velocity 20 m / s at the time 2 s .

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? SOLUTION Height, s = 20m Acceleartion, a = 10m/s2 By 3rd equation of motion v2 = u? + 2as v2 = 0 + 2 x 10 x 20 = 400 = v= 20m/s

Distance covered by the ball, s= 20 m Acceleration, a= 10 m/s 2 Initially, velocity, u= 0 (since the ball was initially at rest) Final velocity of the ball with which it strikes the ground, v According tothe third equation of motion: v 2= u 2+ 2 as v 2= 0 + 2 (10) (20) v= 20 m/s According tothe first equation of motion: v= u+ at Where, Time, ttaken by the ball to strike the ground is, 20 = 0 + 10 ( t) t= 2 s Hence, the ball strikes the ground after 2 s with a velocity of 20 m/s.

Physics A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s², with what velocity will it strike the ground? A ball is gently dropped from a height of 20 m . If its velocity increases uniformly at the rate of 10 m / s² , with what velocity will it strike the ground?

Here, u = 0 , a = 10 m s − 2 , s = 20 m . 2 a s = v 2 − u 2 ∴ 2 × 10 m s − 2 × 20 m = v 2 − 0 ∴ v 2 = 400 m 2 s − 2 t h e r e o e v = 20 m s − 1 v = u + a t ∴ 20 m s − 1 = 0 + 1 0 m s − 2 × t ∴ 20 m s − 1 = 10 m s − 2 × t ∴ t = 20 m s − 1 10 m s − 2 = 2 s The ball will strike the ground with a velocity of 20 m s − 1 . It will strike the ground after 2s.

Dear Student Assume, the final velocity with which the ball will strike the ground be ‘v’and time it takes to strike the ground be ‘t&Rsquo; Initial Velocity of ball, u =0 Distance or height of fall, s =20 m Downward acceleration, a =10 m s^-2 As we know, 2as =v^2-u^2 v^2= 2as+ u^2 = 2 x 10 x 20 + 0 = 400 ∴Final velocity of ball, v = 20 ms^-1 t = (v-u)/a ∴Time taken by the ball to strike = (20-0)/10 = 20/10 = 2 seconds Thanks