Answer:
The height of the hill is, h = 38.42 m
Explanation:
Given,
The horizontal velocity of the soccer ball, Vx = 15 m/s
The range of the soccer ball, s = 42 m
The projectile projected from a height is given by the formula
S = Vx [Vy + √(Vy² + 2gh)] / g
Therefore,
h = S²g/2Vx² (Since Vy = 0)
Substituting the values
h = 42² x 9.8/ (2 x 15²)
= 38.42 m
Hence, the height of the hill is, h = 38.42 m
Answer:
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Explanation:
Answer:
Vo = 4.5 [m/s]
Explanation:
In order to solve this problem, we must use the following equation of kinematics.

where:
Vf = final velocity = 12 [m/s]
Vo = initial velocity [m/s]
a = acceleration = 1.5 [m/s²]
t = time = 5 [s]
Now replacing:
![12=v_{o}+1.5*5\\v_{o}=12- (7.5)\\v_{o}= 4.5[m/s]](https://tex.z-dn.net/?f=12%3Dv_%7Bo%7D%2B1.5%2A5%5C%5Cv_%7Bo%7D%3D12-%20%287.5%29%5C%5Cv_%7Bo%7D%3D%204.5%5Bm%2Fs%5D)
<span>Weight of block, Wb = mass*gravity = 50*9.8 = 490 N</span>
Since block is being pulled up by a 13-degree slope
Therefore, Force which is acting parallel to the slop:
<span> F p =490 Sin </span>
= 110.2N
Force which is acting perpendicular to the slope:
<span> Fv =490 Cos</span>
= 477.4 N
Net force can be given as follows:
<span>F n = (250 - 110.2 - 0.2*</span>477.4) N
<span>Fn=44.3N</span>
Now acceleration is given by the ratio of force to mass
<span>a = Fn/m</span>
<span>=44.3/50 = 0.89 ms^<span>-2</span></span>