You launch a projectile from level ground at a speed of 25.0 m/sm/s and an angle of 36.9∘∘ above the horizontal. How long after it is launched does the projectile reach its maximum height above the ground?

The projectile takes 2.04 s to reach its maximum height.

Explanation:

The maximum height of the projectile is given by the relation :

     .....(1)

Here H is height, u is initial speed, g is acceleration due to gravity and θ is projectile angle with respect to horizontal.

Substitute 25 m/s for u, 9.8 m/s² for g and 36.9° for θ in equation (1).

H = 11.50 m

The equation of motion for vertical direction is :

v = u cosθ - gt

Here v is the vertical speed and t is time.

At maximum height, v is zero. So, the above equation in terms of t becomes :

t = (u cosθ)/g

Substitute the suitable values in the above equation:

t =

t = 2.04 s

the answer is b

explanation:

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂