Question

Is the math on this correct? When I try and do it using all sorts of calculators im coming up with different answers.

2. David kicks a ball at an angle of 53% above the horizontal. Its initial velocity maximum height the ball will reach, the horizontal displacement and total time required for this motion. (sin53°=).8 and cos53°=0.6) 0. Xu X Lovat a safe | Vy X = Vox +dary ILO Ms Cos53 O -38ay = IOYSins3° aZA ) LX E 9.m) A = 3BD E

Answer 1

First of all we will calculate the time required for this motion
t = 2VSin53 /g = 2*10Sin53 /9.81 = 1.63 s
Where V is the velocity of ball by which it was kicked
Now the horizontal distance
= Velocity*time = (10Cos53)*1.63 = 9.799 m
Now for maximum height
= V²Sin² $\theta$ /2g
= 10²Sin²53 /(2*9.81) = 3.25 m
Hence the maximum height will be 3.25 m