S= ut-
gt² is a formula which gives the distance S in meters traveled by a ball from the thrower's hands 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 9.8 m/s²
(i) if a ball is thrown upwards at 14 m/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
Answered by
2
Actually the formula is
s = ut + (1/2)at²
but, as the motion is against gravity, a = - g
∴s = ut - (1/2)gt²
(i)height at 1 sec, u = 14 m/s
substituting the value,
s = 14×1 - (1/2)×9.8×1²
= 14 - 4.9
=9.1 m
(ii) If we take u = 14 m/s, height = 5 m
5 = 14t - (1/2)×9.8t²
⇒4.9t² -14t +5 =0
solving this quadratic equation, we will get two solutions
t = (10+5√2)/7 sec or (10-5√2)/7 sec
(iii) For height 5 m, there are two possible values of time,amazed??
Actually, when the ball is going upward then it reaches first 5 m, the time will be (10-5√2)/7 sec and we also know that, due to law of gravitation, that ball will fall back after reaching the top point, so again it will reaches the 5 m height when falling back, that time will be (10+5√2)/7 sec.
s = ut + (1/2)at²
but, as the motion is against gravity, a = - g
∴s = ut - (1/2)gt²
(i)height at 1 sec, u = 14 m/s
substituting the value,
s = 14×1 - (1/2)×9.8×1²
= 14 - 4.9
=9.1 m
(ii) If we take u = 14 m/s, height = 5 m
5 = 14t - (1/2)×9.8t²
⇒4.9t² -14t +5 =0
solving this quadratic equation, we will get two solutions
t = (10+5√2)/7 sec or (10-5√2)/7 sec
(iii) For height 5 m, there are two possible values of time,amazed??
Actually, when the ball is going upward then it reaches first 5 m, the time will be (10-5√2)/7 sec and we also know that, due to law of gravitation, that ball will fall back after reaching the top point, so again it will reaches the 5 m height when falling back, that time will be (10+5√2)/7 sec.
Answered by
0
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 andt.
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.
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 andt.
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.
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