Home > Physics > Falling Objects

If a projectile is shot at an angle of #(pi)/8# and at a velocity of #14 m/s#, when will it reach its maximum height?

Answer 1

Nora Arzate

#t=0.55 " "s#

#"it can be calculated using t="(v_i*sin alpha)/g#

#v_i=14" "m/s#

#alpha=pi/8#

#g=9.81 N/(kg)#

#t=(14*0.383)/(9.81)#

#t=0.55 " "s#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Xavier Adame

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

[ t_{max} = \frac{v_0 \sin(\theta)}{g} ]

Where:

( t_{max} ) is the time to reach maximum height.
( v_0 ) is the initial velocity (14 m/s in this case).
( \theta ) is the launch angle (( \frac{\pi}{8} ) in radians).
( g ) is the acceleration due to gravity (approximately ( 9.8 , \text{m/s}^2 )).

Plugging in the given values:

[ t_{max} = \frac{14 , \text{m/s} \times \sin\left(\frac{\pi}{8}\right)}{9.8 , \text{m/s}^2} ]

[ t_{max} \approx 0.896 , \text{s} ]

So, the projectile will reach its maximum height approximately 0.896 seconds after being launched.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.