Question

A ball is launched at an angle of 39.8 degrees up from the horizontal, with a muzzle velocity of 6.6 meters per second, from a launch point which is 1 meters above the floor. How high will the ball be above the floor (in meters), when it is a horizontal distance of 2.7 meters away? Use 9.82 meters per second for "g".

Answer 1

Answer:

1.85 m

Explanation:

The horizontal velocity of the ball is

$v_x = v cos \theta = (6.6 m/s) cos 39.8^{\circ}=5.1 m/s$

The horizontal distance travelled is

d = 2.7 m

And since the motion along the horizontal direction is a uniform motion, the time taken is

$t= \frac{d}{v_x}=\frac{2.7 m}{5.1 m/s}=0.53 s$

The vertical position of the ball is given by

$y= h + u_y t - \frac{1}{2}gt^2$

where

h = 1 m is the initial heigth

$u_y = v sin \theta = (6.6 m/s) sin 39.8^{\circ}=4.2 m/s$ is the initial vertical velocity

g = 9.82 m/s^2 is the acceleration due to gravity

Substituting t = 0.53 s, we find the height of the ball at this time:

$y=1 m + (4.2 m/s)(0.53 s) - \frac{1}{2}(9.82 m/s^2)(0.53 s)^2=1.85 m$