Question

a projectile is launched from ground level at an angle of 63.0 with an initial speed of 30.0 m/s. how long does it take to reach max heighg? and what height does it attain?

what is the velocity when the projectile has a height of 28.5 m and is still going up?

Answer 1

Answer 2

Answer:

1. Time to Reach Maximum Height:

$$t=\frac{v_0\sin \theta }{g}=\frac{30.0\sin 63.0\mathrm{°}}{9.81}\approx 2.73\text{s}$$

2. Maximum Height Attained:

$$h_{\text{max}}=\frac{(v_0\sin \theta {)}^2}{2g}=\frac{(30.0\sin 63.0\mathrm{°}{)}^2}{2\times 9.81}\approx 36.5\text{m}$$

3. Velocity at 28.5 m (While Ascending):

• Vertical velocity component ($v_y$) at height $y=28.5\text{m}$:

  $$v_y=\sqrt{(v_0\sin \theta {)}^2-2gy}=\sqrt{(30.0\sin 63.0\mathrm{°}{)}^2-2\times 9.81\times 28.5}\approx 8.57\text{m/s}$$

• Horizontal velocity ($v_x$) remains constant:

  $$v_x=v_0\cos \theta =30.0\cos 63.0\mathrm{°}\approx 13.6\text{m/s}$$

• Resultant velocity:

  $$v=\sqrt{v_x^2+v_y^2}=\sqrt{13.6^2+8.57^2}\approx 16.1\text{m/s}$$

Key Equations Used:

• Time to max height: $t=\frac{v_0\sin \theta }{g}$

• Max height: $h_{\text{max}}=\frac{(v_0\sin \theta {)}^2}{2g}$

• Velocity at height $y$: $v_y=\sqrt{(v_0\sin \theta {)}^2-2gy}$

Assumptions:

• Neglect air resistance.

• $g=9.81{\text{m/s}}^2$