s=ut-1/2gt^2 is a formula which gives the distance s in meters travelled by a ball from the thrower's hand if it is thrown upwards with an initial velocity of u m/s after a time of t seconds. "g" is the acceleration due to gravity and is 9.8m/s^2 (i)If a ball if thrown upwards at 14m/s,how high has it gone after 1 second? (ii)how long does it take for the ball to reach a height of 5 meters? (iii)why are there two possible times to reach a height of 5 meters?
Answers
We will use the give n formula to perform the calculations :
g = 9.8
S = ut - 1/2gt²
i) u = 14 m/s
t = 1
S = 14 × 1 - 0.5 × 9.8 × 1²
= 9.1 m
ii) S = 5
g = 9.8
u = 14
5 = 14t - 0.5 × 9.8t²
5 = 14t - 4.9t²
4.9t² - 14t + 5 = 0
By quadratic formula :
t = {14 +/-√[(-14)²- 4× 4.9 × 5] }/ 2 × 4.9
t = 2.44 or 0.4185
Answer:
ii) using the formula s=ut-0.5gt^2,we obtain two time which are approximately 0.4184s and 2.438. You might be curious why we've got (a negative in "g" instead of positive) it's because "g" opposes motion.
iii)We've got 2 time in which each has different meaning.
The first time(smaller) which is 0.4184s implies the time it takes for the ball to reach 5m at first point of throw.
While the second time(bigger) means the time it takes to get to 5m at first point of throw then to the max height then back to 5m when coming down. Therefore its obvious that the time to reach the max height is less than the second time(bigger) gotten .