a stone is thrown vertically upwards with an initial velocity of 20 metre per second .find the maximum height it reaches and the time taken by it to reaches and the time taken by it to reach the height.
Answers
hope it helps...........
A ball is thrown vertically upward with a speed of 20 m/s. When will it reach the maximum height? What is the maximum height reached?
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.