An object is thrown at an angle of 30 degrees above the horizontal, with a speed of 88.2 m/s.

Question

How long does it take the ball to reach the maximum height?

What is the maximum height reached by the ball?

What is the magnitude or the object's velocity 2 seconds after launch?

What is the horizontal distance covered by the object when it reaches at the highest point?

At what time the speed of the object is 50m/s?

Answer 1

Angle = 30 degree = Theta

V = 88.2 m/s

Horizontal component of initial velocity = 88.2 Cos theta = 88.2 Cos 30 = 88.2 * 88.2 *0.866 = 76.38m/s

Vertical component of initial velocity = 88.2 Sin 30 = 44.1m/s

For maximum height

0 = (44.1) ² - 2 g* Y

Y = Maximum height = 44.1*44.1 / 2*9.8 = 1944.81/ 2*9.8 = 99.26 m

Time for maximum height = t = 44.1/9.8 = 4.5 sec

Horizontal distance covered = Horizontal velocity * time = 76.38*4.5 = 343.71 m

Horizontal speed after 2 sec = 76.38m/s

Vertical speed after 2 sec

= 44.1 - 9.8 *2 = 24.5 m/s

Resultant after 2s = sqrt ( 24.5*24.5 + 76.38*76.38 ) = sqrt (600.25 + 5833.9)= sqrt ( 6434.15) = 80.21 m/s