Question

a projectile is shot upward from the surface of the earth with an initial velocity of 120 meters per second. what is the velocity after 5 seconds? after 10 seconds?

Answer 1

\(\bf h = -9.8t^2+v_ot+h_o \text{ , in meters}\\
v_o = \textit{initial velocity}= 120\\
h_o = \textit{initial height, from the ground} = 0
t = \textit{seconds}
\\\quad \\\quad \\
h = -9.8t^2+v_ot+h_o \implies y = -9.8t^2+120t+0\\\quad \\
\textit{to find it at say 5seconds, set } x = 5seconds\\\quad \\
y = -9.8t^2+120t+0 \implies y = -9.8(5)^2+120(5)+0\\ \implies y = -245+600 \implies y = 355meters\\\quad \\
\textit{at 5seconds, the velocity will be } \cfrac{355m}{5s}
\implies 71\frac{meters}{secs}\)

Answer 2

\(\bf h = -9.8t^2+v_ot+h_o \text{ , in meters}\\
v_o = \textit{initial velocity}= 120\\
h_o = \textit{initial height, from the ground} = 0\\
t = \textit{seconds}\)

Answer 3

try to find the velocity, using the initial velocity equation, for 10seconds :)

Answer 4

@jdoe0001 I don't know why, but my prof taught me that \[x = x_0 + v_0 t+ \frac{1}{2 }at^2\] and if it apply for throwing up a ball, then a = -9.8. while yours is \[x = x_0 + v_0t + at^2\]

Answer 5

One more thing, if we calculate the velocity at t = 5 and t =10, why don't we use v = v_0 + at directly?

Answer 6

hmm... well..... my understanding is that -9.8 is just the gravity pull in meters

Answer 7

I am talking about the formula.

Answer 8

well... that's where the formula gets it from

Answer 9

I understand that you use x to replace to v by that ( I am sorry, not right) formula . My question is the velocity which respect to t is v =v_0 + at helps us find out v (5) and v (10). why do we have to go around like that?

Answer 10

hmm, I see what you mean... finding "x" from the given value of "y"
yes, I could have gone route and solve for "t"

well.... in my case.... usually just to show procedure

Answer 11

\(\large h = -9.8t^2+v_ot+h_o \)

second derivative of this does not equal -9.8. so it needs 1/2 also for t^2 coeffecient

Answer 12

hmm

Answer 13

ok.... seems I do need the 1/2 :/

Answer 14

\(\bf h = -4.9t^2+v_ot+h_o \text{ , in meters}\\
v_o = \textit{initial velocity}= 120\\
h_o = \textit{initial height, from the ground} = 0
t = \textit{seconds}
\\\quad \\\quad \\
h = -4.9t^2+v_ot+h_o \implies y = -4.9t^2+120t+0\\\quad \\
\textit{to find it at say 5seconds, set } x = 5seconds\\\quad \\
y = -4.9t^2+120t+0 \implies y = -4.9(5)^2+120(5)+0\\ \implies y = -122.5+600 \implies y = 477.5meters\\\quad \\
\textit{at 5seconds, the velocity will be } \cfrac{477.5m}{5s}
\implies 95.5\frac{meters}{secs}\)

Answer 15

95.5 is average speed, its not velocity at t=5

Answer 16

to find velocity at t=5, i think we need to do dh/dt, and substitute initial velocity and solv...

v = dh/dt = -9.8t + v0


v at t = 5 = -9.8(5) + 120