If a ball is launched upward at 20 m/s, after 1.5 seconds how high about the point of release in the ball be?
a. 15 m
b. 5 m
c. 18.75 m
d. 20 m
Answers
Answer:
These questions can be answered by making use of Newton's equations of motion. There are 3 equations of motions.
v=u+at
s=ut+12at2
v2=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/s , i.e., u=20m/s , the final velocity that the ball can achieve at the maximum height is 0m/s , hence, v=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/s2 . But for simplicity, we can take the value of a to be 10m/s2 , so a=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 .
Using the first equation,
v=u+at
0=20−10t
10t=20
t=2
Using the third equation,
v2=u2+2as
02=202+2×(−10)×s
20s=400
s=20
Hence, the ball will travel for 2 seconds and complete a distance of 20 metres upwards.
ʜᴏʟᴀ ᴍᴀᴛᴇ !
⭐ᴀɴsᴡᴇʀ⭐
The ball will travel for 2 seconds and complete a distance of 20 metres upwards.