A b and c its so simple bro send it 23-89
Answer:
d_2 = 4d_1
Explanation:
The range or horizontal distance covered by a projectile projected with a velocity U at an angel of θ to the horizontal is given by
R = U²sin2θ/g
Let the range or horizontal distance of ball 1 with initial velocity U projected at an angle θ = 55° be
d_1 = U²sin2θ/g
Let the range or horizontal distance of ball 2 with initial velocity V = 2U projected at an angle θ = 55° be
d_2 = V²sin2θ/g
= (2U)²sin2θ/g
= 4U²sin2θ/g
= 4d_1 (since d_1 = U²sin2θ/g)
So, the ball 2 lands a distance d_2 = 4d_1 from the initial point.
P=w/t
w=15
t=3
therefore, 5 watts (b)
Answer:
176.58 m
Explanation:
t = Time taken = 6 seconds
u = Initial velocity = 0
v = Final velocity
s = Displacement
g = Acceleration due to gravity = 9.81 m/s² = a
Equation of motion
The object travels 176.58 m from the cliff in 6 seconds.
