What was its initial speed?

Question

A ball is thrown straight up from ground level. After a time 5.1 s, it passes a height of 89.3 m. The acceleration due to gravity is 9.8 m/s2. Answer in units of m/s.

Answer 1

Let initial velocity be v_i , time is t=5.1 sec.  Height y is 89.3 m.

y=v_it -1/2 g t²

89.3 = v_i(5.1) - 1/2 (9.8) (5.1)²

89.3 = 5.1 v_i -127.45

5.1 v_i =89.3+127.45

5.1 v_i = 216.75

v_i = 216.75/5.1 = 42.5 m/s