A ball is thrown upward at 12m/s. What is its speed 1.0s later?

Answer 1

Gravity slows down the ball at 9.8m/s/s. After 1s, its velocity is 12-9.8=2.2m/s

Answer 2

using the formula:

1/2GT^2+ViT+Hi
where: G is the constant -9.8m/s, Vi is the initial velocity (12m/s), T is the elapsed time(1.0s), and Hi is the intial height which is 0.

so 1/2(-9.8m/s)(1.0s)^2+12(1.0^s)+0

=7.1m, so after 1 sec of being thrown upwards, the ball is 7.1m from the ground.

and the speed is 2.2m/s

good luck with you assignment.

Answer 3

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Answer 4

v= u+ at

where v is velocity at time t ( 1 second),
u is initial velocity (12 m/s).

a stands for accelaration, and is negative in sign because it acts opposit to direction of motion( towards centre of earth)

so a = -9.8 m/(s^2) , accelaration due to gravity

v=12 - (9.8*1)

=> v = 2.2 m /second, upward direction. But I see you asked only for speed, not direction.

Answer 5

The equation is:
V=V0+at
where:
V0=12m/s
a=-9.80665m/s^2
t=1s
V=12-9.80665(1)=2.19335m/s upward.

Answer 6

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Answer 7

11.4m/s
by using the formula
v^2=u^2-2gs
v^2=144-9.8*2
v^2=124.4
v=11.4m/s

where 9.8 is the gravity and neagtive sign is due to against the gravity

Answer 8

Maybe, but I'm not 100%

Answer 9

2.2m/s (g=9.8m/s squared aproximatelly)

Answer 10

2.2m/s