a ball is thrown upwards with a velocity of 10 m/s.find the maximum hieght reached by the ball aand the time taken to come back to the thrower .( g=10 m/sq
Answers
Answer:
Step-by-step express’s answer using equations for constant accelerations is the way you’ll be taught to do it, and rightly so, but just for fun, here’s another approach.
The ball is thrown with potential energy zero and kinetic energy given by 12mv2. It climbs until kinetic energy is zero and potential energy is given by mgh. We ignore friction, so total energy is conserved
mgh=12mv2
h=v22g
substituting v=10ms−1,g=9.8ms−2
h=1002(9.8)m
h=5.1mlanation:
Answer:
given: u = 10m/sec , g = -10m/sec², v = 0
to find s and total time t
v² - u² = 2gs
0² - 10² = 2×-10×s
therefore s = -100/-20
=> 5 m
time taken to reach max height
s = ut + 1/2 g t²
5 = 10t + 1/2(-10)t²
5 = 10t - 5t²
5t² - 10t + 5 = 0
t² - 2t + 1 = 0
t² - t - t - 1 = 0
t(t - 1) - 1(t - 1) = 0
(t - 1)² = 0
t - 1 = 0
t = 1 sec
OR
total time taken to hit the ground = 2t = 2 secs
v = u + at
0 = 10 + (-10)t
0 = 10 - 10t
0 = 10(1 - t)
t = 1 sec
total time taken to hit the ground = 2t = 2 secs