Answer:
The ball shall keep rising tills its velocity becomes zero. Let it rise to a height h feet from point of projection.
Step-by-step explanation:
Let us take the point of projection of the ball as origin of the coordinate system, the upward direction as positive and down direction as negative.
Initial velocity u with which the ball is projected upwards = + 120 ft/s
Uniform acceleration a acting on the ball is to acceleration due to gravity = - 32 ft/s²
The ball shall keep rising tills its velocity becomes zero. Let it rise to a height h feet from point of projection.
Using the formula:
v² - u² = 2 a h,
where
u = initial velocity of the ball = +120 ft/s
v = final velocity of the ball at the highest point = 0 ft/s
a = uniform acceleration acting on the ball = -32 ft/s²
h = height attained
Substituting the values we get;
0² - 120² = 2 × (- 32) h
=> h = 120²/2 × 32 = 225 feet
The height of the ball from the ground at its highest point = 225 feet + 12 feet = 237 feet.
Pretty sure it’s 25 but I could be wrong. It’s also rly blurry. Hope this helps
E to F = 678.35 miles
F to G = 156.8 miles
G to H = x
Total distance = 2,457 miles
E to G = 678.35 + 156.8 miles
E to G = 835.15
G to H = 2,457 - 835.15
G to H = 1621.85 miles
Answer:110
Step-by-step explanation:x^2+17x=15x+35
x^2+17x-15x-35=0
x^2+2x-35=0
delta=2^2-4*1*(-35)=4+140=144
x1=(-2+V144)/2=(-2+12)/2=10/2
x=5
so 15*5+35=75+35=110
