Question

A projectile is fired up into the air at a speed of 165 m/s at an angle of 75° relative to the horizontal. Ignore air drag. A.) Determine the MAXIMUM HEIGHT (above where it was thrown) the projectile will reach. B.) Determine how LONG the projectile will be in the air. Assume it lands at the same height it was launched. C.)Assuming it lands at the same height it was launched at, determine how FAR the projectile will land from where it was launched. D.) Assuming it lands at the same height it was launched at, determine how FAST the projectile will be going upon impact. E.) Assuming it lands at the same height it was launched at, determine the ANGLE (relative to the horizontal) of impact. °

Answer 1

A) 1296.3 m

The initial velocity of the projectile is

$u = 165 m/s$

and the angle is

$\theta=75^{\circ}$

So, we can find the initial vertical velocity of the projectile, which is given by

$u_y = u sin \theta = (165 m/s)sin 75^{\circ}=159.4 m/s$

The motion of the projectile along the vertical axis is a uniformly accelerated motion, with constant acceleration

$g=-9.8 m/s^2$

where the negative sign means the direction is towards the ground. The maximum height is reached when the vertical velocity becomes zero: therefore, we can use the following SUVAT equation

$v_y ^2 - u_y^2 = 2gh$

where

$v_y = 0$ is the final vertical velocity

h is the maximum height

Solving for h, we find

$h=-\frac{-u_y^2}{2g}=\frac{-(159.4 m/s)^2}{2(-9.8 m/s^2)}=1296.3 m$

B) 32.5 s

In order to determine how long the projectile will be in the air, we need to find the time t at which the projectile reaches the ground.

We can find it by analyzing the vertical motion only. The vertical position at time t is given by

$y(t) = u_y t + \frac{1}{2}gt^2$

By substituting y(t) = 0, we find the time at which the projectile reaches the ground. We have:

$0= u_y t + \frac{1}{2}gt^2\\0 = t(u_y + \frac{1}{2}gt)$

which has two solutions:

t = 0 --> beginning of the motion

$u_y + \frac{1}{2}gt=0\\t=-\frac{2u_y}{g}=-\frac{2(159.4 m/s)}{-9.8 m/s^2}=32.5 s$

C) 1387.8 m

The range of the projectile can be found by analyzing the horizontal motion only.

In fact, the projectile travels along the horizontal direction by uniform motion, with constant horizontal velocity, given by:

$u_x = u cos \theta = (165 m/s) cos 75^{\circ}=42.7 m/s$

So, the horizontal position at time t is given by

$x(t) = u_x t$

If we substitute

t = 32.5 s

which is the time at which the projectile reaches the ground, we can find the total horizontal distance covered by the projectile.

So we have:

$x= u_x t = (42.7 m/s)(32.5 s)=1387.8 m$

D) 165 m/s

The speed of the projectile consists of two independent components:

- The horizontal velocity, which is constant during the motion, and which is equal to

$v_x = u_x = 42.7 m/s$

- The vertical velocity, which changes during the motion, given by

$v_y = u_y + gt$

Substituting

$u_y = 159.4 m/s$

and

$t=32.5 s$

We find the vertical velocity when the projectile reaches the ground

$v_y = 159.4 m/s + (-9.8 m/s^2)(32.5 s)=-159.4 m/s$

which is the same as the initial vertical velocity, but with opposite direction.

Now that we have the two components, we can calculate the actual speed of the projectile:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(42.7 m/s)^2+(-159.4 m/s)^2}=165 m/s$

and the final speed is exactly equal to the initial speed, since according to the conservation of energy, the projectile has lost no energy during the motion.

E) $-75^{\circ}$

The angle of impact is given by

$\theta = tan^{-1} (\frac{|v_y|}{v_x})$

where

$|v_y| = 159.4 m/s$ is the final vertical velocity

$v_x = 42.7 m/s$ is the final horizontal velocity

We have taken the absolute value of $v_y$ since $v_y$ is negative; this means that the resulting angle will be BELOW the horizontal. So we have:

$\theta = tan^{-1} (\frac{159.4 m/s}{42.7 m/s})=75^{\circ}$

which means $-75.0^{\circ}$, below the horizontal.