tangential acceleration without circular motion?

[Godhelpme]

A body jumps with initial speed 10m/s under  a 30 degree angle with the horizon. Air resistence shouldn't be taken into account and g=10m/s2. What will the tangential acceleration be after 0.5 seconds??

Edited October 16, 2019 by Godhelpme

[swansont]

3 minutes ago, Godhelpme said:

A body jumps with initial speed 10m/s. Air resistence shouldn't be taken into account and g=10m/s2. What will the tangential acceleration be after 0.5 seconds??

You need to get/provide more information. Tangential to what?

[studiot]

I think you are talking about the projectile equations under gravity, but where is your attempt?

Edited October 16, 2019 by studiot
spelling

[Country Boy]

The acceleration due to gravity is -g (which you are taking to be -10) vertically.  Taking y to be positive upward and x positive to the right, the equations of motion are x= v_x t  and y= v_y t- (g/2)t^2.  The initial speed is 10 m/s at 30 degrees so v_x= 10(cos(30))= 10(1/2)= 5 m/s and v_y= 10(sin(30))= 10(sqrt(3)/2)= 5 sqrt(3) m/s which is about 8.7 m/s.  So we have x= 5t and y= 8.7t- 5t^2.  From x= 5t, t= x/5 so y= 8.7(x/5)- 5(x^2/25)= 1.74x- 0.2x^2.  That is, of course, a parabola opening downward.  The slope of the tangent line, at a given x, is 1.74- 0.4x.  After 0.5 seconds, x= 5(0.5)= 2.5 and y= 8.7(0.5)- 5(0.25)= 5.6.  The slope of the tangent line is 1.74- 0.4(2.5)= 0.74 so the tangent line to the trajectory at that time is y= 0.74(x- 2.5)+ 5.6.  The tangential acceleration is the projection of the vector (0, -10) on that line.