Projectiles - how high is the ball

[Huskies213]

Can anyone help with these 2?

I can't seem to figure out how to get the answer of 2.8m for this ...
IF you throw a ball at an initial velocity of 26 m/s at an angle of 20 above the horizontal, how high above the projection point is the ball after 2.8 s?

and also

You drag a crat weighing 311 N across a floor by pulling on a rope attatched to the crate. You exert a force of 450N on the rope inclined at 38 degrees, and the floor exerts a horizontal force of 125 N, the acceleration is ...
(do you add the forces, for total force? what do you do with the 38 degrees ?)

 

[siddharth]

For the first one, what have you done? Have you split the components of velocity along the x-axis and y-axis?

For the second one, have you learned about the resolution of forces along the x and y directions.

If you have a book, try reading some example problems. This is very basic stuff.

 

[kyle8921]

For the first question, the height of the ball after 2.8 seconds can be calculated using the formula h = h0 + v0t + (1/2)at^2, where h0 is the initial height (in this case, 0m), v0 is the initial velocity (26 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time (2.8 s). Plugging in these values, we get h = 0 + (26)(sin20)(2.8) + (1/2)(-9.8)(2.8)^2 = 2.8m. This means that after 2.8 seconds, the ball will be 2.8 meters above the projection point.

For the second question, the total force can be calculated using the formula F = ma, where m is the mass of the crate (311N/9.8m/s^2 = 31.73kg) and a is the acceleration. The acceleration can be found by breaking the forces into their components. The horizontal force of 125N will not contribute to the acceleration since it is perpendicular to the direction of motion. The vertical component of the 450N force is 450sin38 = 275.6N, which will contribute to the acceleration. Using the formula F = ma, we get 275.6N = (31.73kg)a, and solving for a, we get a = 8.69 m/s^2. So the acceleration of the crate is 8.69 m/s^2.