<span>2.4854847</span> miles per hour
Answer:
28.5 m/s
18.22 m/s
Explanation:
h = 20 m, R = 20 m, theta = 53 degree
Let the speed of throwing is u and the speed with which it strikes the ground is v.
Horizontal distance, R = horizontal velocity x time
Let t be the time taken
20 = u Cos 53 x t
u t = 20/0.6 = 33.33 ..... (1)
Now use second equation of motion in vertical direction
h = u Sin 53 t - 1/2 g t^2
20 = 33.33 x 0.8 - 4.9 t^2 (ut = 33.33 from equation 1)
t = 1.17 s
Put in equation (1)
u = 33.33 / 1.17 = 28.5 m/s
Let v be the velocity just before striking the ground
vx = u Cos 53 = 28.5 x 0.6 = 17.15 m/s
vy = uSin 53 - 9.8 x 1.17
vy = 28.5 x 0.8 - 16.66
vy = 6.14 m/s
v^2 = vx^2 + vy^2 = 17.15^2 + 6.14^2
v = 18.22 m/s
Answer:
3.416 m/s
Explanation:
Given that:
mass of cannonball 
= 72.0 kg
mass of performer 
= 65.0 kg
The horizontal component of the ball initially 
= 6.50 m/s
the final velocity of the combined system v = ????
By applying the linear momentum of conservation:
v = 3.416 m/s
Answer:
x is vertical and y is horizontal
Explanation:
The pressure exerted by the concrete cylinder is 2.60 pound/in².
We need to know about the pressure to solve this problem. Pressure is a unit that describes how much force is applied to a surface area. It can be determined as
P = F / A
where P is pressure, F is force and A is area.
From the question above, we know that
F = 375 pound
A = 144 in²
By substituting the given parameters, we can calculate the pressure
P = F / A
P = 375 / 144
P = 2.60 pound/in²
Thus, the pressure should be 2.60 pound/in².
Find more on pressure at: brainly.com/question/25965960
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