2nd and only 2nd option is right
Answer:
The speed of the ambulance is 4.30 m/s
Explanation:
Given:
Frequency of the ambulance, f = 1790 Hz
Frequency at the cyclist, f' = 1780 Hz
Speed of the cyclist, v₀ = 2.36 m/s
let the velocity of the ambulance be 'vₓ'
Now,
the Doppler effect is given as:

where, v is the speed of sound
since the ambulance is moving towards the cyclist. thus, the sign will be positive
thus,

on substituting the values, we get

or
vₓ = 4.30 m/s
Hence, <u>the speed of the ambulance is 4.30 m/s</u>
The velocity increased from 4 m/s to 22 m/s in 3 seconds. 18 m/s in 3 seconds so the average acceleration is change in velocity divided by time. 18 m/s divided by 3 seconds = 6 m/s^2
Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding