An object thrown vertically upwards from the top
of a tower of height 39.2 m, reaches the ground in
4s. The velocity with which it is thrown upwards is
(a) 9.8m/s
(b) 4.9m/s
(c) 19.6m/s
(d) 39.2m/s
I need full explanation.
Answers
Answer:
(a) 9.2 m/s
Explanation:
Assuming the only force acting on the ball is the initial velocity and standard gravity (9.81 m/s^2): the ball’s position can be written as y=(0.5*a*(t^2))+(v*t)+y1, where ‘a’ is gravity, ‘v’ is the velocity the object is thrown with, ‘y1’ is the initial height in meters, ‘t’ is the time in seconds, and ‘y’ is the height of the object in meters. Using an up as positive reference and plugging in givens, the equation becomes 0=(0.5*(-9.81 m/s^2)*((4 sec)^2))+(v*(4 sec))+(39.2 m). Solving for ‘v’ gives 9.82 m/s..
Formula,
s = ut - [1/2]gt^2
4u -8g= - 39.2
u -2g=-9.8
u = 9.8 m/s
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Answer:
Explanation:
s = ut - 1/2gt^2
39.2 = 4u - 1/2(g)(16)
39.2 = 4u - 8g
4(u - 2g) = 39.2
u - 2g = 9.8
u = 9.8 + 2g
u = 9.8 + 2(-9.8)
u = 9.8 -19.6
u = -9.8 m/s
Since its the velocity in the upward direction its taken positive i.e. 9.8m/s