If a projectile is shot at an angle of #pi/2# and at a velocity of #34 m/s#, when will it reach its maximum height??

Question

If a projectile is shot at an angle of #pi/2# and at a velocity of #34   m/s#, when will it reach its maximum height??

Charles Aeschliman · Accepted Answer

The upward motion of the projectile will be due to its vertical component of velocity,now initially during the projection,its vertical component of velocity was #34 sin (pi/2) =34ms^-1# So,if it reaches its maximum height at time #t# at that instant its upward component of velocity will be zero, so,we can use, #v=u-g t# Thus, we can write here: #0=34 -g t# so, #t=3.47 s# Well look carefully here the projectile was thrown directly upwards,i.e it had no horizontal displacement,as its horizontal component of velocity was #34 cos (pi/2) =0#

Carson Alexander · Answer

undefined The projectile will reach its maximum height when its vertical velocity component becomes zero. Using the equation for vertical motion:

v_f = v_i + at

Where:
v_f = final velocity (0 m/s at maximum height)
v_i = initial velocity in the vertical direction (0 m/s)
a = acceleration due to gravity (-9.8 m/s²)
t = time

Substituting the given values:

0 = 34 * sin(π/2) + (-9.8) * t

Solving for t:

t = (34 * sin(π/2)) / 9.8
t ≈ (34 * 1) / 9.8
t ≈ 3.47 seconds

Therefore, the projectile will reach its maximum height after approximately 3.47 seconds.