the height x at time t of a particle projected vertically upwards from the ground is given by x(t)=32t-16t^2?

Question

find
a) the initial velocity
b)the greatest height reached
c) the total time taken to return to the ground

Answer 1

Hi,

a) the initial velocity 32
The initial velocity is the value of b in at² + bt + c.

b)the greatest height reached 16
The axis of symmetry is x = -b/(2a) = -32/(2*-16) = 1
this means the highest point (vertex) occurs when x = 1. When x = 1, y = 16, the maximum height

c) the total time taken to return to the ground 2
If it took one second to reach the vertex it takes another to return to the ground. This can also be found by factoring:
32t - 16t² = 0
16t(2 - t) = 0
16t = 0 and 2 - t = 0 solve to t = 0 and t = 2 as the 2 times the particle is on the ground.

I hope that helps!! :-)

Answer 2

A) initial velocity: derive x(t) and evaluate at t=0, you get 32.

B) greatest height is the maximum of 32t-16t^2, so the derivative of x(t) (the velocity) must be zero there:

dx/dt = 32 - 32 t

it's 0 for t=1 and the greatest height is x(1)=32 - 16 = 16

C) It's t=2 because x(2)=0. Also from B) we can say that
the particle takes one more second to go back to ground,
hence 1+1=2.

Answer 3

x(t)= -16t² + 32t

Take the first derivative to get velocity:
x'(t) = -32t + 32
x'(0) = 32
I assume height is in feet and t is in seconds, but since the problem didn't give units, I can't say 32 ft/sec.

Initial velocity = 32 height units per time unit
=======================
Velocity = 0 at the instant the particle reaches greatest height.
0 = -32t + 32
t = 1

x(1) = -16·1² + 32·1
= -16 + 32
= 32 height units
=========================
height = 0 when particle returns to ground.
0 = -16t² + 32t
= 16t² - 32t
= t² - 2t
= t(t-2)
t = 0, t = 2

particle returns to ground in 2 time units

Answer 4

a) Initial velocity

Differentiate the expression : 32-32t and sub t=0

Answer : 32 [m/s]

b) find when dx/dt = 32 - 32t = 0 (when t=1). Differentiate it again to find out whether it is a maximum. Remember, when the second differential is negative, it is a maximum. The subtitute t=1 into the original equation (Answer =16 [m] )

c) Find the value of t when x=0 (except when t=0) (Answer: t=2)

Answer 5

V = dx/dt = 32-32t
Vo = 32 - 32(0) = 32

greatest height = 32*1 -16*1^2 = 16

2 seconds because x = 0 when t = 2

Answer 6

Vo= dx/dt at t=0 Which is
32-32t=32
Greatest height reached is when V(t) is zero. Since it is fired upwardly, it gradually gets slower and slower due to gravity and finally stops and starts to fall back down
V(t)=dx/dt=32-32t so at t=1 second, it reaches its max height.
x(1)=32-16=16
It takes the same time to rise to max height as it does to fall back to the ground so its flight time is 2 seconds

Answer 7

a) s=vt-(1/2)at^2=32t-16t^2 => vt-(1/2)at^2=32t-16t^2 =>
initial velocity v=32 and acceleration a=32

b) To reach the top must 32 t1=16 t1^2 => 16 t1=32 =>
t1=2 seconds

c) Total time is t=t1+t2 => t=2+t2
h=16t1^2=64 =(1/2)gt2^2 => (1/2).9.81.t2^2=64 =>
4.905 t2^2=64 => t2=3.61 sec and t=2+3.61=5.61 sec

Answer 8

The equation z(t)= -(one million/two)g(t^two) + Ut is right however no longer too useful on this case! Instead, you can use v^two = u^two+2as, in which v = zero, u= U, a = -nine.eight = -g and s = ? This offers zero = U^two -2gs, so 2gs = U^two giving s = U^two/2g. s refers to distance, as a result top. Hope that is transparent!