Question

9. A ball is dropped from a height of 50m. If its velocity increases uniformly at the
rate of 10m/sec2
a) With what velocity will it strike the ground?
b) After what time will it
strike the ground?​

Answer 1

Answer:

vo=0

$v = \sqrt{vo + 2gh} \\ v = \sqrt{0 + 2 \times 10 \times50 \\ } \\ v = \sqrt{1000} \\ = 10 \sqrt{10} m/s \\ \\ t = v - vo \div g \\ t = 10 \sqrt{10} \div 10 \ \\ = \sqrt{10} s$

Answer 2

Here displacement =50m(given)
Acceleration = 10m/sec2(given)
Initial velocity =0, because the ball is dropped from a height, that is from top to bottom
Whenever a substance is dropped from top to bottom, the initial velocity will be zero and final velocity will be some value, but whenever a substance is thrown from bottom to top, the initial velocity be some value and final velocity will be equals to zero as it stops after some time and again comes to ground.
Final velocity = v
v^2-u^2=2as
v2-(0)2=2(10)(50)
v2=1000
V=10root10

b)s=ut+1/2at^2
50=0(t)+1/2(50)(t)^2
50=25t^2
t^2=25
t=5sec




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