Question

A ball is thrown vertically upwards with velocity 50m/s at t=0. Find the distance travelled from t=0 to t=8s

Answer 1

Answer:

80m

Explanation:

Velocity of the ball = u=50m/s  (Given)

Time = t = 0 (Given)

Distance travelled from t = 0 to t = 8 - ?

a = -10

According to the second equation of motion -

s = ut+1/2at²

Thus, substituting the value -

s = 50 × 0 + 1/2 × (-10) × 0

s = 0

Now, again,u=50m/s , a = -10, and  t=8sec

 s = ut+1/2at²

Thus, substituting the value -

s = 50 × 8 + 1/2 × -10 × 8²

s = 400-320

s = 80

Thus, the distance travelled from t = 0 to t = 8s is 80m.

Answer 2

Answer:

170 metrs

Explanation:

First find time when body has max height:

T=u/g

T=5 sec

So body travels max height till t = 5sec then returs to ground for 3 sec.

For t=5,

H=(u^2)/2g [ since v at top =0 ]

H1= 125 metres

For last 3 sec,

u =0

H=1/2gt^2

H2= 45 meters

Total distance in 8 sec= H1+H2

=125+45 =170m