use a(t)=-32ft/s^2 as the accleration due to gravity. a ball is thrown vertically upward from the ground with?

Question

use a(t)=-32ft/s^2 as the acceleration due to gravity. a ball is thrown vertically upward from the ground with an intial velocity of 56 per second.for how many seconds will the ball be going up ward?

please show th work

Answer 1

EXACTLY 1 3/4 seconds, i.e. 1.75 seconds (given your input data).

The upward speed v = 56 ft/s + a t ; since a = - 32ft/s^2, v = o when t = 56/32 s. Dividing by the common factor of 8 in the numerator and denominator, this is 7/4 s or 1 3/4 seconds.

Note that it really isn't desirable to write 'a' as a(t). The latter form implies that 'a' is a function of time. But the whole point about problems involving gravity near the Earth is that 'a' (i.e.'g') is a CONSTANT. And indeed, your value "a(t)=-32ft/s^2" IS a constant; it exhibits NO dependence on time. (It's that many ft/s^2 no matter how much time has elapsed.)

Live long and prosper.

Answer 2

V=56-32t=0
32t=56
t=56/32=1.75 sec

Answer 3

Well v = at, so 56 ft/s - (32 ft/s²)(t sec) = 0 solves out at 56 = 32t, t = 1.75 s.

Just to double check it, h = vt - 16t²
On the ground, h = 0, so
0 = 56t - 16t²
0 = t(7 - 2t)
t = 0 or t = 3.5

Since half the time the ball is going up and half the time it's going down (symmetry of the parabola), it's at the top of its flight at half of 3.5 = 1.75 s.

Answer 4

Approximately 1.76 seconds. It took me a while to figured it out because I'm use to the standard velocity in meters of 9.8 meters per second.