Two seconds after being projected from ground level, a projectile?

Question

Two seconds after being projected from ground level, a projectile
is displaced 40 m horizontally and 53 m vertically above its launch point. What are the
(a) horizontal and (b) vertical components of the initial velocity of the projectile? (c) At
the instant the projectile achieves its maximum height above ground level, how far is it
displaced horizontally from the launch point?

Answer 1

Vh = 20 m/s
Vv = 36.3 m/s

Now you solve the rest of it ☺


Doug

Answer 2

To find the horizontal and vertical components of the initial velocity, use the equation:

[change in displacement]/[change in time]

(a) Horizontal component: v_o = 40/2 = 20 m/s (no air resistance in horizontal direction)

(b) Vertical component: Use x = v_0*t + 0.5*a*t^2
Plug in t = 2, x =53, and solve for v_0

v_0 = 53-(0.5*9.8*2^2)/2 = 16.7 m/s

(c) To find the time where the projectile reaches its peak, you first need to find the intial velocity of the projectile and at what angle the projectile is shot.

To find the total initial velocity, use the Pythagorean Theorem:

v_o(total) = sqrt(v_0[in x-direction]^2 + v_0[in y-direction]^2)

= sqrt(20^2 + 16.7^2) = 26.06 m/s = v_0(total)

Now you need to find the angle the projectile is show in order to find the final displacement. You need to use SOHCAHTOA. You can use any relationship in SOHCAHTOA to find the angle, but using the x and y components will give you a more accurate angle.

Therefore angle = inverse tangent (y/x) = inverse tangent (16.7/20) = 39.86(degrees)

Now you have all the components to solve (c)

To find the time at which the projectile will reach maximum, use the equation:

v_f = at+v_0. Solving for 't' we get, t = [v_f-v_0]/a

For this part, we only use the vertical component of v_0 because 'a' = gravity only moves in the vertical direction.

Therefore, t= [0-16.7]/(-9.8) = 1.7 seconds

Now you can find the horizontal displacement when the projecticle reaches the peak

'x' = v_o(x-drection)*t+0.5at^2 (a = 0 since gravity does not move in the horizontal direction)

Therefore 'x' = 20*1.7 = 34 meters

Hope this helps

Answer 3

Let x and y be the horizontal and vertical components at the time of projection.
a]vertical component
S=Ut+1/2*g*t^2
53=y*2+1/2*9,81*2^2
2y=53-19.62=23.38
y=11.69m/sec
b]horizontal component
In this direction there is no deceleration
assuming no air resistance
2x=40m
x=20met/sec
c] Assuming no air resistance the trajectory
will be a symmetrical parabola
from symmetry
horiz. displacement at the time
it achieves it peak=1/2*40m=20m