Answer :
Answer:
car land from base at distance 39.66 m
impact speed = 45.17 m/s
Explanation:
given data
speed v = 20 m/s
height h = 28 m
angle θ = 20°
to find out
How far from the base of cliff does the car land and car impact speed
solution
we consider here x is horizontal direction and
and y is vertical direction
so we find time first for climb 28 m high
use equation of motion for climb height
- h = v×sinθ × t - 0.5× a×t² ............1
here h is height, t is time , a is acceleration and v is speed
put here all value
-28 = 20×sin20 × t - 0.5× 9.8 ×t²
t = 4.89 seconds
so
distance = ut = 20×cos20 × t = 20×cos20 × 4.86
distance = 39.66 m
so car land from base at distance 39.66 m
and
for impact speed we calculate vx and vy
we use equation of motion
v = u - at ..................2
at vertical direction
vy = 20×sin20 - 9.8 ( 4.89) = -41.08 m/s
at horizontal direction
vx = 20×cos20
vx = 18.79 m/s
so impact speed = [tex]\sqrt{vx^{2} + vy^{2}}[/tex] .................3
put here value
impact speed = [tex]\sqrt{18.79^{2} + (-41.08)^{2}}[/tex]
impact speed = 45.17 m/s
The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.
- a) The distance of the base of the cliff where the car land is 39.66 m.
- b) The car's impact speed is 45.17 m/s.
What is the equation of motion?
The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.
The second equation of the motion for distance can be given as,
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
Here, u is the initial body, a is the acceleration of the body and t is the time taken by it.
Given information-
The speed of the car is 20 m/s.
The height of the cliff is 28 meters.
The cliff is inclined upward at an angle of 20 degrees.
- a)The distance of the base of the cliff where the car land-
As the initial speed of the car in the vertical direction can be given with the sine angle. Thus put the values in the above formula as,
[tex]28=-20\sin (20^o)+\dfrac{1}{2}\times9.8t^2\\t=4.89\rm s[/tex]
Now the distance traveled by the object is the product of speed and the time. Thus,
[tex]d=20\cos(20^o)\times4.86\\d=39.66\rm m[/tex]
Hence the distance of the base of the cliff where the car land is 39.66 m.
- b) The car's impact speed-
Let [tex]v_x[/tex] is the velocity of the car in horizontal direction and [tex]v_y[/tex] is the velocity of the car in vertical direction. Thus the impact speed can be given as,
[tex]v_i=\sqrt{v_x^2+v_y^2}\\v_i=\sqrt{20\cos(20)^2+(20\sin(20^o)-9.8\times4.89)^2}\\v_i=45.17\rm m/s[/tex]
Hence, the car's impact speed is 45.17 m/s.
Thus,
- a) The distance of the base of the cliff where the car land is 39.66 m.
- b) The car's impact speed is 45.17 m/s.
Learn more about the equation of motion here;
https://brainly.com/question/13763238