Question

A ball is thrown vertically upward from the ground with a velocity of 20m/s (a)How long will it take to rise to its highest point? (b) How high does the ball rise? (c) How long after projection will the ball have a velocity of 10 m/s upward?

Answer 1

Answer:

Time to reach maximum height can be obtained from v=u+at

0=20+(−10)t

t=2s

s=ut+0.5at

2

=20(2)+0.5(−10)(2)

2

=20m

Thus, total distance for maximum height is 45 m

s=ut+0.5at

2

45=0+0.5(10)(t

′

)

2

t

′

=3s

Total time= 3+2= 5s

Answer 2

Answer:

a) 2.04 s

b)20.408m

c)1.02 s

Explanation:

a) V = U + at

terminal velocity, V is zero

U is initial velocity, ie 20m/s

a is same as g but in opposite direction, so a = -9.8

so, 0 = 20 + (-9.8)t

t = 20/9.8 = 2.04s

b) H = U^2÷2g

H = 20^2 ÷ 2×9.8

H = 400/19.6 = 20.408m

c) Here V is 10 m/s

so, From V = U + at

10 = 20 + (-9.8)t

t = 10/9.8 = 1.02s