Question #c6d6a

Your projectile is fired with an initial velocity of 22 m/s and at an angle of #33^@# with the horizontal.

#v_0 = "22 m/s"#

#theta=33^@#

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This means that the motion of the projectile has a vertical component and a horizontal component.

Vertically, the projectile will be acted upon by the gravitational acceleration, #g#. This means that you can write

#d_y = v_(0y) * t - 1/2 * g * t^2#, where

#d_y# - the total vertical displacement;
#v_(0y)# - the vertical component of #v_0#;
#t# - the total time of flight.

In your case #d_y# will be equal to zero because the projectile starts from the ground level and ends up on the ground level. The vertical component of #v_0# will be equal to

#v_(0y) = v_0 * sintheta#

The equation becomes

#0 = v_0 * (sintheta) * t - 1/2 * g * t^2#

#-1/2 * 9.8t^2 + 22 sin(33^@)t = 0#

#-4.9t^2 + 11.98t = 0 <=> t(-4.9t + 11.98) = 0#

The only acceptable solution is

#t = (-11.98)/-4.9 = color(green)("2.44 s")#

At the maximum vertical height, the vertical component of the velocity is zero. This means that you can write

#0 = v_(0y)^2 - 2 * g * h#, where

#h# - the maximum height the projectile reaches.

Solving this equation for #h# will get you

#h = (v_0^2sin(33^@)^2)/g = color(green)("7.32 m")#

Horizontally, the projectile is not acted upon by any forces. This means that you can write

#d_x = v_(0x) * t#, where

#d_x# - the range of the projectile;
#v_(ox)# - the horizontal component of #v_0#.

Use the values you have to solve for #d_x#

#d_x = v_0 * costheta * t = 22"m"/cancel("s") * cos(33^@) * 2.44cancel("s")#

#d_x = color(green)("45.0 m")#

it is defined as the time taken by any object to travel in any medium it is calculated by

t=2vsintheta/a
maximum vertical height is the peak of that projectile motion the object is travelling
horizontal range it is the range of horizontal motion which is done by object