<span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal 
it can cross and reach top of trajectory if its top height h = 1.5m 
and horizontal distance d = (1/2) Range 
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let t be top height time 
at top height, vertical component of its velocity =0 
vy = 0 = u sin p - gt 
t = u sin p/g 
h = [u sin p]*t - 0.5 g[t[^2 
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g 
u^2 sin^2 p/2g = 1.5 
u^2 sin^2 p = 1.5*2*9.8 = 29.4 
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component 
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t = HALF the time of flight 
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g] 
1 = u^2 sin p cos p/g 
u sin p * u cos p = 9.8 
5.42 * u cos p = 9.8 
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component 
check>> 
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s 
u < less than fish's potential jump speed 6.26 m/s 
so it will able to cross</span>
        
             
        
        
        
Answer:
ρ = 7500 kg/m³
Explanation:
Given that
mass ,m = 12 kg
Displace volume ,V= 1.6 L
We know that
1000 m ³ = 1 L
Therefore V= 0.0016 m ³
When metal piece is fully submerged
We know that 
mass = Density x volume

Now by putting the values in the above equation

ρ = 7500 kg/m³
Therefore the density of the metal piece will be  7500 kg/m³.
 
        
             
        
        
        
Uhh it is used to detirmine heat
        
                    
             
        
        
        
Answer:
hmax=81ft
Explanation:
Maximum height of the object is the highest vertical position along its trajectory.
The vertical velocity is equal to 0 (Vy = 0)

we isolate th (needed to reach the maximum height hmax)

The formula describing vertical distance is:

So, given y = hmax and t = th, we can join those two equations together:


if we launch a projectile from some initial height h all you need to do is add this initial elevation


 
        
             
        
        
        
Answer:
given , v = 300 km/hr; distance d = 1500 km; then time t = d/v = 1500/300 = 5 hrs
Explanation: