Question

A toy projectile is fired from the ground vertically upward with an initial velocity of +29 meters per second. The projectile arrives at its maximum altitude in 3.0 seconds. [Neglect air resistance.] 1. The greatest height the projectile reaches is approximately

Answer 1

Answer:

43m

Explanation:

According to projectile, the maximum height reached by the toy is expressed as:

H = u²sin²Ф/2g

u is the initial velocity = 29m/s

Ф = 90° (projected vertically upward)

g is the acceleration die to gravity = 9.81m/s²

Substitute into the formula:

H = u²sin²Ф/2g

H = 29²sin²90/2(9.81)

H = 841(1)/19.62

H = 42.86m

Hence the greatest height the projectile reaches is approximately 43m

Answer 2

The greatest height the toy projectile reaches =

$H_{max} = 42.86\; m$

Projectile motion is an example  a plane motion in two dimensions.

The equations for projectile motion are derived from the equations of motions applied in x and y directions.

The maximum height of the flight is achieved at the point where the velocity in  +y  direction becomes Zero and just after that the projectile keeps accelerating.

The maximum height of the projectile is given by equation (1)

$H_{max} = \dfrac{u^2 Sin^2\theta }{2g} .....(1)\\H_{max} = Maximum\; Height\; of\; the\; Projectile \\\theta = Angle \; of \; projection \; in \; degree\; from \; Horizontal\; axis \\u = Initial\; velocity\; of \; the \; projectile\\ g = Acceleration \; due \; to \; gravity = 9.81 \; m/s^2$

Initial velocity = 29 m/s

Angle of projection = 90 ° (vertically upwards)

Putting in equation (1)

$H_{max} = (29)^2 \times Sin^2 90\textdegree/(2\times 9.81)$

$H_{max} = 42.86\; m$

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https://brainly.com/question/21241023