Question

Q2) An object is thrown vertically upwards with an initial velocity of
100m/s. Find a) the time taken to reach the maximum height.
b) maximum height reached.
c) velocity after 10s. (take g= 10m/s2

Answer 1

Step-by-step explanation:

u = 100

At maximum height v = 0

s = maximum height

acceleration a = - 10

3rd equation of motion

v ^2 = u^2 + 2as

0 = 100^2 + 2(-10)s

s = 10000/20

s = 500 meter

To find time taken (t):

By first equation of motion

v = u + at

0 = 100 + (-10) t

t = 10 s

To find velocity after 5 s (v):

v = u + at

v = 100 + (-10)5

v = 50m/s

Answer 2

Answer:

Step-by-step explanation:

u = 100 m/s

At maximum height v = 0

s = maximum height

acceleration a = - 10

3rd equation of motion

v ^2 = u^2 + 2as

0 = 100^2 + 2(-10)s

s = 10000/20

s = 500 meter

To find time taken (t):

By first equation of motion

v = u + at

0 = 100 + (-10) t

t = 10 s

To find velocity after 5 s (v):

v = u + at

v = 100 + (-10)5

v = 50m/s