A ball is thrown vertically downwards with a velocity of 20m/s

Time taken by first ball to reach the ground is #T = sqrt((2H)/g) = sqrt(("2 × 20.0 m")/("10 m/s"^2)) = "2 s"# Time taken by second ball to reach the ground #= "2 s" - "1 s" = "1 s"# For second ball #S = ut + 1/2at^2# #"20.0 m" = (u × "1 s") + (1/2 × "10 m/s"^2 × ("1 s")^2)# #"20 = u + 5"# #"u = 15 m/s"# [ #color(red)("Note:")# I took #"g = 10 m/s"^2# instead of #"9.8 m/s"^2# to make the calculations easier.]

Method I: We will breah the motion in two parts. O to A (vertically upward motion): At the highest point, velocity=0 v = u − g t 0 = 20 − 10 t 1 ⇒ t 1 = 2 s v 2 = u 2 − 2 g h 0 = ( 20 ) 2 − 2 × 10 × h ⇒ h = 20 m A to B (vertically downward direction): A B = h + 60 = 20 + 60 = 80 m 80 = 0 + 1 2 g t 2 2 = 5 t 2 2 ⇒ t 2 = 4 s The ball will strike the ground after 6 s. v 0 = u + g t 2 = 0 + 10 × 4 = 40 m / s Note: The above method is lengthy. We will not break motion in part unless necessary. We know that equations of motion are applicable for displacement. Method II: Now taking O as the origin and vertically upward direction positive, from O to B through A, the displacement is negative, i.e. h = − 60 m h = u t − 1 2 g t 2 ( u ↑ ⏐ g ⏐ ↓ ) − 60 ≡ 20 t − 5 t 2 t 2 − 4 t − 12 = 0 ( t − 6 ) ( t + 2 ) = 0 t = 6 s , t = − 2 s Time t = − 2 s is not possible, time cannot be-ve. The ball will steike the ground after 6 s. v = u − g t v 0 = 20 − 10 × 6 = − 40 m / s Negative sign indicates that the velocity is in downward direction.

The correct option is D 20 m/s When it hits the ground T E = 1 2 m v 2 0 + m g h It loses 50% energy, it left with 50% energy, this left energy converted into potential energy 50 100 [ 1 2 m v 2 0 + m g h ] = m g h m v 2 0 2 + m g h = 2 m g h m v 2 0 + 2 m g h = 4 m g h v 2 0 = 2 g h v 2 0 = 2 × 10 × 20 v 2 0 = 400 v 0 = 20 m / s

A ball is thrown down vertically with an initial speed of 20 m/s from a height of 60 m. Find (a) its speed just before it strikes the ground and (b) how long it takes for the ball to reach the ground. Repeat (a) and (b) for the ball thrown directly up from the same height and with the same initial speed. Take g = 10 m/s². Transcribed Image Text: 6. A ball is thrown down vertically with an initial speed of 20 m/s from a height of 60 m. Find (a) its speed just before it strikes the ground and (b) how long it takes for the ball to reach the ground. Repeat (a) and (b) for the ball thrown directly up from the same height and with the same initial speed. Take g = 10 m/s?. A skier leaves the end of a horizontal ski jump at 22.0 m/s and falls through a vertical distance of 3.20 m before landing. Neglecting air resistance, (a) how long does it take the skier to reach the ground? (b) How far horizontally docs the skier travel in the air before landing? (Sec Section 3.4.) A projectile is thrown from the top of a building with an initial velocity of 30 m/s in thehorizontal direction. If the top of the building is 30 m above the ground, a) how long isthe projectile in the air? b) With what speed and angle will the projectile be moving justbefore it strikes the ground? (Hint: final speed = 38.6 m/s) "An object is shot straight up from the ground with an initial velocity of 20m/s. How high above the ground is the object one second after it reaches its maximum height? Neglect air resistan...

When the stone is thrown vertically upwards, #u# = initial velocity = # 20#m/s It will reach some height where its final velocity will become zero, #v = 0#m/s #a= -g = -10ms^(-2)# ----negative as it is going in direction against the gravitational pull. (Consider #g= 10 m/s^2# for ease of calculation). Height = distance it travels upwards will be : #s = (v^2 -u^2)/ (2a)# # s = (0 - 20^2)/ (-2xx10)# m #s = -400/-20 #m #s = 20#m Time taken to reach this distance will be : # t = (v-u)/a # # t = (0- 20)/-10 = 2# # t = 2# seconds ----------let this be #t_1# So it reaches #20#m from the peak of tower of height #25#m. So to reach to ground,the stone has to cover: #s= 20+25= 45#m. When it starts falling from #45#m above ground, its initial velocity is zero, so: #u = 0# We know it will come down with uniform #a=g =10 ms^-2# We know : #s = ut + 1/2 at^2# # 45 = 0xxt + 1/2 xx10xxt^2# # 45xx2/10 = t^2# # cancel45^9xx2/cancel10^2 = t^2# # 9xx cancel2/cancel2 = t^2# #therefore t^2 = 9#, i.e. # t = 3# seconds-----Let this be #t_2# Total time taken after throwing the stone upwards will be : #t = t_1+ t_2# #t = 2+3= 5#seconds #therefore#The stone will reach the ground in 5 seconds after being thrown upwards from the tower.