00:01
Hi there, here we have to solve an assignment of four questions and in the first we have to find the total time of the flight of the ball.
00:16
Let's do this.
00:18
So just to remind, peter throws the ball with an initial speed of 35 .6 meters per second at an angle of 26 .7 degree above the horizontal.
00:34
And the ball is released from the initial height of 15 meters.
00:43
So let's first illustrate this system.
00:48
Let's introduce y and x -axis.
00:51
The ball is released from the initial height of h -0 and it abase the parabolic trajectory.
01:01
So it reaches the maximum height somewhere here.
01:06
And so this initial speed of v0 makes an angle of alpha with the horizontal so therefore y component total time of light is dictated by y component of the speed and the initial y component of the speed equals to v0 times sine alpha overall vy as a function of time equals to v0 sine alpha minus g t and total time of the flight equals is the sum of two components, t1 and t2.
01:47
So t1 is the time of moving from initial height to maximum height.
01:59
And this time t1 equals to v0, sine alpha, divided by g.
02:05
Let's calculate it.
02:07
That is 35 .6 meters per second.
02:10
Times sign of 26 .7 degree divided by 9 .80 meters per second squared.
02:41
That equals to 3 .63 seconds.
02:46
The time t2 is time needed to go from maximum height to the ground.
02:53
And so the maximum height then equals to g, t2 squared, t2 squared over 2.
03:03
C2 equals to square root of 2h maximum over g we need to find this maximum height and so to do this we'll calculate yeah we'll calculate that so h maximum equals to the initial speed sorry to the initial height plus height achieved due to the vertical component of the speed that's what which is v0 sine alpha squared divided by 2g...