A ball is projected vertically upwards from the ground with speed 21 ms^–1. The ball moves freely under gravity once projected. What is the greatest height reached by the ball?

Set out information given in question, and taking the upward direction to be positive: s (displacement) = ?, u (initial speed) = 21ms-1, v (final speed at maximum height) = 0ms-1, a (acceleration when falling freely under gravity) = - 9.8ms-2, t = ?. Using v2 = u2 + 2as: 0 = 212 + 2(-9.8)s, s = 441 / 19.6 = 22.5m. So maximum height reached = 22.5m

SS

Related Maths A Level answers

All answers ▸

What is the derivative of f(x)=sqrt(3x+2)=(3x+2)^(1/2)?


Let N be an integer not divisible by 3. Prove N^2 = 3a + 1, where a is an integer


integral of (tan(x))dx using the substitution u = cos(x)


How to do the product rule for differentiation