If a projectile is shot at an angle of #(7pi)/12# and at a velocity of #9 m/s#, when will it reach its maximum height?

Question

If a projectile is shot at an angle of #(7pi)/12# and at a velocity of #9  m/s#, when will it reach its maximum height?

Layla Ahmed · Accepted Answer

#0.63 sec#

We know that according to the law of motion . #V=U+a*t#

#:.# this law is true for the frame of reference for non inertial frame of reference .

So it can be written as , #V_y=U_y+a_y*t#

we know that vertical component of the velocity is #Usin theta# here it can be written as

#Usin((7pi)/12)=Usin(pi-(3pi)/12)=Usin(pi/4)###as sin is positive in 2 quadrant

for attaining maximum height we get final velocity #V_y=0# so upon substituting the values in the equation we get . #0=9*1/sqrt(2)-10t#

as the acceleration due to gravity is in negative direction. so we get .

#9/(10*sqrt(2))=t# #t=(9sqrt(2))/20=12.69/2=0.63 sec#

Benjamin Asiedu · Answer

undefined To find the time it takes for the projectile to reach its maximum height, we can use the formula:

$$ t = \frac{v_0 \sin(\theta)}{g} $$

Where:
- $ t $ = time taken to reach maximum height
- $ v_0 $ = initial velocity (9 m/s)
- $ \theta $ = launch angle ($ \frac{7\pi}{12} $)
- $ g $ = acceleration due to gravity (9.8 m/s²)

Substituting the given values:

$$ t = \frac{9 \times \sin\left(\frac{7\pi}{12}\right)}{9.8} $$

$$ t ≈ \frac{9 \times 0.707}{9.8} $$

$$ t ≈ \frac{6.363}{9.8} $$

$$ t ≈ 0.65 \, \text{seconds} $$

Therefore, the projectile will reach its maximum height after approximately 0.65 seconds.