A stone was thrown upwards with an initial velocity of 20.0 m/s frọm a height of 50.0 m Find the following: a. the time required to reach the maximum height, (1) vys = Vyi – gt (2) 0= 20- (9.8)t tmaxh = 2.04 s b. the maximum height that can be reached by the stone, (1) hmax = Ys = yi + 0.5(vi + v)t (2) hma = 50 + 0.5(20 + 0)(2.04) hmax = 70.4 m C. the velocity of the stone when it returns to the height from which it was thrown, and d. the time for the stone to reach the ground
A stone was thrown upwards with an initial velocity of 20.0 m/s frọm a height of 50.0 m Find the following: a. the time required to reach the maximum height, (1) vys = Vyi – gt (2) 0= 20- (9.8)t tmaxh = 2.04 s b. the maximum height that can be reached by the stone, (1) hmax = Ys = yi + 0.5(vi + v)t (2) hma = 50 + 0.5(20 + 0)(2.04) hmax = 70.4 m C. the velocity of the stone when it returns to the height from which it was thrown, and d. the time for the stone to reach the ground
Related questions
Question
100%
Answer c and d
Transcribed Image Text:A stone was thrown upwards with an initial velocity of 20.0 m/s from a height of 50.0 m
Find the following:
a.
the time required to reach the maximum height,
(1) vyf = Vyi - gt
(2)
0 = 20 – (9.8)t
tmaxh = 2.04 s
b. the maximum height that can be reached by the stone,
(1) hmax = y1 = yi + 0.5(v; + v)t
(2) hmax = 50 + 0.5(20 + 0)(2.04)
hmax
= 70.4 m
с.
the velocity of the stone when it returns to the height from which it was thrown, and
d. the time for the stone to reach the ground
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 2 images
