Question

A projectile is fired with an initial speed of 100 m/s and angle of elevation 30 degrees. The projectile eventually hits the ground. How far from the initial position (which can be assumed to be the origin) does it travel? Find the speed at impact (with the ground). Note that the projectile is sibject to the downward acceleration due to gravity of 9.8meters per sq. sec

Answer 1

Answer:

Explanation:

Given

launch velocity $u=100\ m/s$

Launch angle $\theta =30^{\circ}$

Range of Projectile $R=\frac{u^2\sin 2\theta }{g}$

$R=\frac{100^2\times \sin (2\times30)}{9.8}$

$R=883.699\approx 883.7$

Horizontal velocity remain same and only vertical velocity changes.

Initially  vertical  velocity is in upward direction but as soon as it reaches the ground its direction change but magnitude remain same.

$u_y=100\sin (30)=50\ m/s$

$u_x=100\cos (30)=86.60\ m/s$

$u_{net}=\sqrt{(50)^2+(86.60)^2}$

$u_{net}=99.99\approx 100\ m/s$