A cannon is fired at 30 degrees from the ground and the cannonball has initial velocity of 15 m/s. What is the height of the highest point the cannonball reaches and how far is this point horizontally from the cannon?
With this type of question always draw a diagram with the values on it. We can assume the positive direction is upwards. Start with considering the vertical motion and use SUVAT. s= ? (this is what we are looking for- the height above the ground of the cannonball). u= 15sin30 = 7.5 m/s (this is the vertical component of the initial velocity of 15 m/s). v= 0 (the vertical component of velocity at the highest point is always zero). a= -g = -9.81 m/s2 (this is acceleration due to gravity- it is negative because it is acting downwards). t is unknown so we use the corresponding suvat equation (without t in it) v2 = u2 + 2as . Rearrange for s and plug in values to find that s= 2.87 m (height at the highest point). Consider the horizontal motion ( where horizontal velocity is constant for projectile motion). distance = speed x time = 15cos30 x t. Find time using v=u+at from vertical values, as vertical and horizontal time are the same. t= 0.76 s . So the horizontal distance of the heighest point is x= 15cos30 x 0.76 = 9.87 m (horizontal distance from highest point).
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