Question

A stone is thrown vertically upwards with an initial velocity of 20 m/sec find the maximun height it reaches and the time taken by it to reach height .(g=10m/s

Answer 1

Answer:

These questions can be answered by making use of Newton's equations of motion. There are 3 equations of motions.

v=u+atv=u+at

s=ut+12at2s=ut+12at2

v2=u2+2asv2=u2+2as

Where,

v = final velocity

u = initial velocity

a = acceleration

t = time

s = distance

In your question, the initial velocity is given as 20m/s20m/s, i.e., u=20m/su=20m/s, the final velocity that the ball can achieve at the maximum height is 0m/s0m/s, hence, v=0m/sv=0m/s. Since the only first that cause the acceleration is gravity, a is taken as g where g is acceleration due to gravity, and had a value of 9.81m/s29.81m/s2. But for simplicity, we can take the value of a to be 10m/s210m/s2, so a=10m/s2a=10m/s2. Now, we need to find, what's s and t.

Note: Since the ball is thrown upwards, which is against the force of gravity (gravity always acts downwards), we need take the value of a (in this case, g) as −10m/s2−10m/s2.

Using the first equation,

v=u+atv=u+at

0=20−10t0=20−10t

10t=2010t=20

t=2t=2

Using the third equation,

v2=u2+2asv2=u2+2as

02=202+2×(−10)×s02=202+2×(−10)×s

20s=40020s=400

s=20s=20

Hence, the ball will travel for 2 seconds and complete a distance of 20 metres upwards.