A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight that's 5 meters above the ground. At what rate
is the tip of his shadow moving? We've already set this up part of the way. We know that dx/dt=-2 meters per secon
1 answer:
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Answer:
Explanation:
Ball is thrown downward:
initial velocity, u = - 20 m/s (downward)
height, h = - 60 m
Acceleration due to gravity, g = - 9.8 m/s^2 (downward)
(a) Let the speed of the ball as it hits the ground is v.
Use third equation of motion
v = 39.69 m/s
(b) Let t be the time taken
Use First equation of motion
v = u + a t
- 39.69 = - 20 - 9.8 t
t = 2 second
Now the ball is thrown upwards:
initial velocity, u = 20 m/s (upward)
height, h = - 60 m
Acceleration due to gravity, g = - 9.8 m/s^2 (downward)
(c) Let the speed of the ball as it hits the ground is v.
Use third equation of motion
v = 39.69 m/s
(d) Let t be the time taken
Use First equation of motion
v = u + a t
- 39.69 = + 20 - 9.8 t
t = 6.09 second
Answer:
Explanation:
As we know that ball travels horizontal distance of 24.7 m with uniform speed 49.4 m/s
so we will have
now in the same time ball is turned by angle
now we know that
now the tangential speed of a point at equator is given as